quadratic equation problem solving

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• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 2 Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation...

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Quadratic equation

• 1 Factoring
• 2 Completing the square
• 3 Quadratic Formula

The purpose of factoring is to turn a general quadratic into a product of binomials . This is easier to illustrate than to describe.

Completing the square

Quadratic Formula

See Quadratic Formula .

• Discriminant
• Vieta's Formulas
• Quadratic Inequality
• Factoring Quadratics
• Quadratic equations

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Solving Quadratics by the Quadratic Formula

How to solve quadratic equations using the quadratic formula.

There are times when we are stuck solving a quadratic equation of the form [latex]a{x^2} + bx + c = 0[/latex] because the trinomial on the left side can’t be factored out easily. It doesn’t mean that the quadratic equation has no solution. At this point, we need to call upon the straightforward approach of the quadratic formula to find the solutions of the quadratic equation or put simply, determine the values of [latex]x[/latex] that can satisfy the equation.

In order use the quadratic formula, the quadratic equation that we are solving must be converted into the “standard form”, otherwise, all subsequent steps will not work. The goal is to transform the quadratic equation such that the quadratic expression is isolated on one side of the equation while the opposite side only contains the number zero, [latex]0[/latex].

Take a look at the diagram below.

In this convenient format, the numerical values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easily identified! Upon knowing those values, we can now substitute them into the quadratic formula then solve for the values of [latex]x[/latex].

The Quadratic Formula

• Where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are the coefficients of an arbitrary quadratic equation in the standard form, [latex]a{x^2} + bx + c = 0[/latex].

Slow down if you need to. Be careful with every step while simplifying the expressions. This is where common mistakes usually happen because students tend to “relax” which results to errors that could have been prevented, such as in the addition, subtraction, multiplication and/or division of real numbers.

Examples of How to Solve Quadratic Equations by the Quadratic Formula

Example 1 : Solve the quadratic equation below using the Quadratic Formula.

By inspection, it’s obvious that the quadratic equation is in the standard form since the right side is just zero while the rest of the terms stay on the left side. In other words, we have something like this

This is great! What we need to do is simply identify the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] then substitute into the quadratic formula.

That’s it! Make it a habit to always check the solved values of [latex]x[/latex] back into the original equation to verify.

Example 2 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation is absolutely not in the form that we want because the right side is NOT zero. I need to eliminate that [latex]7[/latex] on the right side by subtracting both sides by [latex]7[/latex]. That takes care of our problem. After doing so, solve for [latex]x[/latex] as usual.

The final answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = – {2 \over 3}[/latex].

Example 3 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation looks like a “mess”. I have variable [latex]x[/latex]’s and constants on both sides of the equation. If we are faced with something like this, always stick to what we know. Yes, it’s all about the Standard Form. We have to force the right side to be equal to zero. We can do just that in two steps.

I will first subtract both sides by [latex]5x[/latex], and followed by the addition of [latex]8[/latex].

Values we need:

[latex]a = – 1[/latex], [latex]b = – \,8[/latex], and [latex]c = 2[/latex]

Example 4 : Solve the quadratic equation below using the Quadratic Formula.

Well, if you think that Example [latex]3[/latex] is a “mess” then this must be even “messier”. However, you’ll soon realize that they are really very similar.

We first need to perform some cleanup by converting this quadratic equation into standard form. Sounds familiar? Trust me, this problem is not as bad as it looks, as long as we know what to do.

Just to remind you, we want something like this

Therefore, we must do whatever it takes to make the right side of the equation equal to zero. Since we have three terms on the right side, it follows that three steps are required to make it zero.

The solution below starts by adding both sides by [latex]3{x^2}[/latex], followed by subtraction of [latex]3x[/latex], and finally the addition of [latex]5[/latex]. Done!

After making the right side equal to zero, the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easy to identify. Plug those values into the quadratic formula, and simplify to get the final answers!

Example 5 : Solve the quadratic equation below using the Quadratic Formula.

First, we need to rewrite the given quadratic equation in Standard Form, [latex]a{x^2} + bx + c = 0[/latex].

• Eliminate the [latex]{x^2}[/latex] term on the right side.

• Eliminate the [latex]x[/latex] term on the right side.

• Eliminate the constant on the right side.

After getting the correct standard form in the previous step, it’s now time to plug the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] into the quadratic formula to solve for [latex]x[/latex].

• From the converted standard form, extract the required values.

[latex]a = 1[/latex], [latex]b = – \,4[/latex], and [latex]c = – \,14[/latex]

• Then evaluate these values into the quadratic formula.

You may also be interested in these related math lessons or tutorials:

Quadratic Formula Practice Problems with Answers

Solving Quadratic Equations by Square Root Method Solving Quadratic Equations by Factoring Method Solving Quadratic Equations by Completing the Square

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Quadratic Equation Word Problems

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

• By Completing the Square
• By Factoring
• By Quadratic Formula
• By graphing

For each process, follow the following typical steps:

• Make the equation
• Solve for the unknown variable using the appropriate method
• Interpret the result

The product of two consecutive integers is 462. Find the numbers?

Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.

Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9

A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet

The difference between the squares of two consecutive even integers is 68. Find the numbers.

Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18

The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.

Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches

A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².

Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?

Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height  x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

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Quadratic Equation Solver

We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :

Is it Quadratic?

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions.

Learn more at Quadratic Equations

Solving Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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10 Quadratic Equations Word Problems

Quadratic equations word problems are math problems in which the equations are not given directly. These problems can be solved by using the given information to obtain a quadratic equation of the form $latex ax^2+bx+c$. We can then use the factoring method, the completing the square method or the quadratic formula to solve the equation.

Here, we will look at 10 quadratic equations word problems with answers. In addition, you will also be able to practice with 5 word problems to solve.

Relevant for …

Learning to solve word problems of quadratic equations.

See problems

10 quadratic equations word problems with answers

Quadratic equations word problems to solve.

The following problems are solved by forming a quadratic equation with the given word problem. Try to solve the problems yourself before looking at the solution.

If the sum of two numbers is equal to 17 and their product is equal to 60, what are the numbers?

We can use the letters X (smaller number) and Y (larger number) to represent both numbers. Then, from the statement, we can obtain the following equations:

$latex X+Y=17~~[1]$

$latex XY=60~~[2]$

Now, we can rearrange equation 1 to get $latex Y=17-X$. Substituting this expression into the second equation, we have:

$latex X(17-X)=60$

Expanding the parentheses and rearranging the equation, we have:

$latex 17X-X^2=60$

$latex X^2-17X+60=0$

Now, we can factor the equation to solve:

$latex (X-12)(X-5)=0$

$latex X-12=0~~$ or $latex ~~X-5=0$

$latex X=12~~$ or $latex ~~X=5$

Therefore, we have:

• If $latex X=5$, we have $latex Y=17-5=12$. This solution is correct because X<Y.
• If $latex X=12$, we have $latex Y=17-12=5$

The numbers are 12 and 5.

The difference between two numbers is equal to 5 and their product is equal to 126. What are these numbers?

Using the information in the statement and using the letters X (smaller number) and Y (larger number) to represent the numbers, we can obtain the following equations:

$latex Y-X=5 ~~[1]$

$latex XY=126~~[2]$

Now, we can write the first equation as $latex Y=5+X$ and substituting it into the second equation, we have:

$latex X(5+X)=126$

$latex 5X+X^2=126$

$latex X^2+5X-126=0$

Now, we can factor and solve:

$latex (X+13)(X-9)=0$

$latex X+13=0~~$ or $latex ~~X-9=0$

$latex X=-13~~$ or $latex ~~X=9$

• If $latex X=-13$, we have $latex Y=5-13=-8$
• If $latex X=9$, we have $latex Y=5+9=14$. This solution is correct because $latex XY=126$

If the area of a rectangle is 78 square units and its longest side is 7 units longer than its shortest side, what are the lengths of the sides?

We can represent the shorter side with x . This means that the longest side is equal to x +7.

Also, we remember that the area of a rectangle is equal to the product of its sides. Therefore, we can form the following equation:

$latex x(x+7)=78$

Expanding and writing the equation in the form $late ax^2+bx+c=0$, we have:

$latex x^2+7x-78=0$

Now, we can use factorization to solve:

$latex (x+13)(x-6)=0$

$latex x+13=0~~$ or $latex ~~x-6=0$

$latex x=-13~~$ or $latex ~~x=6$

Since we can’t have negative lengths, we have $latex x=6$, so the lengths are 6 and 13.

Find the lengths of the sides of a rectangle that has an area of 200 square units if the longer side is twice the length of the shorter side.

Using the letters X (short side) and Y (long side) to represent the sides, we can obtain the following equations using the given information:

$latex Y=2X~~[1]$

$latex XY=200~~[2]$

Now, we can substitute equation 1 into equation 2 and we have:

$latex X(2X)=200$

$latex 2X^2=200$

To solve this equation, we have to isolate X² completely and take the square root of both sides:

$latex X^2=100$

$latex X=\pm\sqrt{100}$

$latex X=\pm 10$

Since a length can’t be negative, we have $latex X=10$. Therefore, we have $latex Y=20$.

The difference between the squares of two consecutive odd numbers is equal to 48. Find both numbers.

Representing the first number with x , we can deduce that a consecutive odd number is equal to $latex x+2$. Therefore, using the information in the statement, we form an equation with the squares of the consecutive numbers:

$latex (x+2)^2-x^2=48$

Expanding, simplifying and solving, we have:

$latex x^2+4x+4-x^2=48$

$latex 4x+4=48$

$latex 4x=44$

$latex x=11$

The consecutive odd numbers are 11 and 13.

The sides of a right triangle have lengths x , x +2, and 10. If 10 is the hypotenuse of the triangle, find the value of x .

Since the triangle is a right triangle, we can use the Pythagorean theorem to find an equation for the lengths of the three sides of the triangle.

The statement tells us that 10 is the hypotenuse of the triangle. Therefore, we have:

$latex x^2+(x+2)^2=10^2$

Expanding and simplifying the equation, we have:

$latex x^2+x^2+4x+4=100$

$latex 2x^2+4x+4=100$

$latex 2x^2+4x-96=0$

Now, we can divide the entire equation by 2 to simplify it:

$latex x^2+2x-48=0$

Solving by factorization, we have:

$latex (x+8)(x-6)=0$

$latex x+8=0~~$ or $latex ~~x-6=0$

$latex x=-8~~$ or $latex ~~x=6$

Since we can’t have a negative length, the answer is $latex x=6$.

If the product of two numbers is equal to 48 and their average is equal to 7, find both numbers.

We can represent one of the numbers with x . This means that the other number is equal to 48/ x since their product is equal to 48.

Therefore, considering that the average of two numbers is equal to the sum of the numbers divided by 2, we have:

$latex \frac{x+48/x}{2}=7$

Simplifying, we have:

$latex x+\frac{48}{x}=14$

$latex x^2+48=14x$

$latex x^2-14x+48=0$

Now, we can solve by factorization:

$latex (x-8)(x-6)=0$

$latex x-8=0~~$ or $latex ~~x-6=0$

$latex x=8~~$ or $latex ~~x=6$

The numbers are 6 and 8.

If the side length of a square is increased by 4, its area is multiplied by 9. Find the side length of the original square.

In this problem, there are two areas that we need to consider, the area of the original square and the area of the larger square (length +4).

Using the x to represent the sides of the original square and remembering that the area of a square is equal to the length of one of the sides squared, we have:

Larger area = 9 × Original area

$latex (x+4)^2=9x^2$

Expanding and simplifying, we have:

$latex x^2+8x+16=9x^2$

$latex 8x^2-8x-16=0$

We can divide the entire equation by 8 and solving by factoring, we have:

$latex x^2-x-2=0$

$latex (x-2)(x+1)=0$

$latex x-2=0~~$ or $latex ~~x+1=0$

$latex x=2~~$ or $latex ~~x=-1$

Since a length cannot be negative, the answer is $latex x=2$.

The shortest side of a right triangle is 4 units less than its hypotenuse. The difference between the short side and the intermediate side is 2 units. If the shorter side is x -2, find the value of x .

From the statement, we can deduce the following lengths of the triangle:

• Short side = x -2
• Intermediate side = x (2 units difference with short side)
• Hypotenuse = x + 2 (4 units more than short side)

Now, we can use the Pythagorean theorem to write an equation for the three lengths:

$latex (x-2)^2+x^2=(x+2)^2$

$$x^2-4x+4+x^2=x^2+4x+4$$

$$x^2-4x+4+x^2-x^2-4x-4=0$$

$latex x^2-8x=0$

$latex x(x-8)=0$

$latex x=0~~$ or $latex ~~x=8$

Since the length cannot be 0, we have $latex x=8$.

PROBLEM 1 0

The figure below has an area of 100 square units. Find the value of x .

We can divide the figure into two rectangles, one rectangle has the area 7 x and the other rectangle has the area x (2 x +3). Adding these areas, we have:

$latex 7x+x(2x+3)=100$

$latex 7x+2x^2+3x=100$

$latex 2x^2+10x-100=0$

Now, we can divide the entire equation by 2 and solve by factoring:

$latex x^2+5x-50=0$

$latex (x+10)(x-5)=0$

$latex x=-10~~$ or $latex ~~x=5$

Since we can’t have a negative length, the answer is $latex x=5$.

Solve the following problems using any method of solving quadratic equations.

Find two numbers such that their sum equals 18 and their product equals 56.

Choose an answer

Find the length of the sides of a rectangle that has an area of 84 square units if the length of one side is 5 units longer than the other side.

The three sides of a right triangle are x, x+1, and 5. if the hypotenuse equals 5, what is the value of x, the difference of the squares of two consecutive even numbers is 68. find the two numbers., if the length of the sides of a square is increased by 6, its area is multiplied by 16. find the length of one of the sides of the original square..

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The most commonly used methods for solving quadratic equations are:

1 . Factoring method

2 . Solving quadratic equations by completing the square

3 . Using quadratic formula

In the following sections, we'll go over these methods.

Method 1A : Factoring method

If a quadratic trinomial can be factored, this is the best solving method.

We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.

Example 01: Solve $ x^2 \color{red}{-8}x \color{blue}{+ 15} = 0 $ by factoring.

Here we see that the leading coefficient is 1, so the factoring method is our first choice.

To factor this equation, we must find two numbers ( $ a $ and $ b $ ) with a sum is $ a + b = \color{red}{8} $ and a product of $ a \cdot b = \color{blue}{15} $.

After some trials and errors, we see that $ a = 3 $ and $ b = 5 $.

Now we use formula $ x^2 - 8x + 15 = (x - a)(x - b) $ to get factored form:

Divide the factored form into two linear equations to get solutions.

Method 1B : Factoring - special cases

Example 02: Solve $ x^2 -8x = 0 $ by factoring.

In this case, (when the coefficient c = 0 ) we can factor out $ \color{blue}{x} $ out of $ x^2 - 8x $.

Example 03: Solve $ x^2 - 16 = 0 $ by factoring.

In this case, ( when the middle term is equal 0) we can use the difference of squares formula.

Method 3 : Solve using quadratic formula

This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.

Example 05: Solve equation $ 2x^2 + 3x - 2 = 0$ by using quadratic formula.

Step 1 : Read the values of $a$, $b$, and $c$ from the quadratic equation. ( $a$ is the number in front of $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)

Step 2 :Plug the values for a, b, and c into the quadratic formula and simplify.

Step 3 : Solve for $x_1$ and $x_2$

Method 2 : Completing the square

This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.

Example 04: Solve equation $ 2x^2 + 8x - 10= 0$ by completing the square.

Step 1 : Divide the equation by the number in front of the square term.

Step 2 : move $-5$ to the right:

Step 3 : Take half of the x-term coefficient $ \color{blue}{\dfrac{4}{2}} $, square it $ \color{blue}{\left(\dfrac{4}{2} \right)^2} $ and add this value to both sides.

Step 4 : Simplify left and right side.

Step 5 : Write the perfect square on the left.

Step 6 : Take the square root of both sides.

Step 7 : Solve for $x_1$ and $x_2$ .

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The Splitting Characteristic Finite Difference Domain Decomposition Scheme for Solving Time-Fractional MIM Nonlinear Advection–Diffusion Equations

• Published: 01 July 2024
• Volume 100 , article number  49 , ( 2024 )

Cite this article

• Zhongguo Zhou   ORCID: orcid.org/0000-0001-9009-3404 1 ,
• Sihan Zhang 1 &
• Wanshan Li 2  

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In this paper, we develop a new splitting characteristic finite difference scheme for solving the time-fractional mobile-immobile nonlinear advection–diffusion equation by combining non-overlapping block-divided domain decomposition method, the operator splitting technique and the characteristic finite difference method. Over each sub-domain, the solutions and fluxes along x -direction in the interiors of sub-domains are computed by the implicit characteristic finite difference method while the intermediate fluxes on the interfaces of sub-domains are computed by local multi-point weighted average from the approximate solutions at characteristic tracking points which are solved by the quadratic interpolation. Secondly, the solutions and fluxes along y direction in the interiors of sub-domains are computed lastly by the implicit characteristic difference method while the time fractional derivative is approximated by L 1-format and the intermediate fluxes on the interfaces of sub-domains are computed by local multi-point weighted average from the approximate solutions at characteristic tracking points are solved by the quadratic interpolation. Applying Brouwer fixed point theorem, we prove strictly the existence and uniqueness of the proposed scheme. The conditional stability and convergence with \(O\left( {\varDelta t}+{\varDelta t}^{2-\alpha }+{h}^2+{H}^\frac{5}{2}\right) \) of the proposed scheme are given as well. Numerical experiments verify the theoretical results.

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The characteristic difference DDM for solving the time-fractional order convection–diffusion equations

An efficient conservative splitting characteristic difference method for solving 2-d space-fractional advection–diffusion equations

The conservative characteristic difference method and analysis for solving two-sided space-fractional advection-diffusion equations

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Acknowledgements

We would like to thank two reviewers’ comments and suggestions on our manuscript which have helped to improve the paper greatly.

This work was supported by Natural Science Foundation of Shandong Government (Grant Nos. ZR2021MA002, ZR2022MF239), and National Natural Science Foundation of China (Grant No. 11701282).

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School of Information Science and Engineering, Shandong Agricultural University, Taian, 271018, Shandong, China

Zhongguo Zhou & Sihan Zhang

School of Mathematics and Statistics, Nanjing University of Science and Technology, Nanjing, Jiangsu, 210094, China

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Zhou, Z., Zhang, S. & Li, W. The Splitting Characteristic Finite Difference Domain Decomposition Scheme for Solving Time-Fractional MIM Nonlinear Advection–Diffusion Equations. J Sci Comput 100 , 49 (2024). https://doi.org/10.1007/s10915-024-02603-4

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Received : 28 September 2023

Revised : 31 May 2024

Accepted : 17 June 2024

Published : 01 July 2024

DOI : https://doi.org/10.1007/s10915-024-02603-4

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