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The calculator on this page shows how the quadratic formula operates, but if you have access to a graphing calculator you should be able to solve quadratic equations, even ones with imaginary solutions.

• Step 1) Most graphing calculators like the TI- 83 and others allow you to set the "Mode" to "a + bi" (Just click on 'mode' and select 'a+bi').
• If you can set your calculator's mode to a + bi you should be able to even calculate imaginary solutions.
• The rest of the steps just involve typing in a,b and c. Make sure that you divide the entire numerator by 2a, just use parentheses.
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How to Solve Quadratic Equations with Complex Numbers

When solving quadratic equations using the quadratic formula , you sometimes get a negative value under the square root. In these cases, the equation does not have any real solutions.

But now that you’re working with complex numbers , you’re able to find all the solutions to quadratic equations. The reason for this is the fact that the imaginary unit i can be utilized to find complex solutions to the quadratic formula.

Complex Quadratic Equations

Let a , b , c ∈ ℂ be complex numbers with a ≠ 0 . Then a z 2 + b z + c = 0 has the following solutions:

If the expression b 2 − 4 a c is negative, you have to use the imaginary unit i to find the solutions.

Solve z 2 − 4 z + 5 = 0 for z

You recognize the coefficients of the equation to be a = 1 , b = − 4 and c = 5 . This means that you can use the quadratic formula to find the solutions:

Since you have a negative number inside the square root, the equation has no real solutions. However, by utilizing the imaginary, you can still find complex solutions:

Thus the final solutions to the equation are:

Note! In Example 1 you used the following relation:

This relation does not hold in general for complex numbers. Using this relation for complex numbers can yield inconsistencies. The following is an example of such inconsistencies:

The reason this calculation yields a contradiction is that − 1 ⋅ − 1 ≠ − 1 ⋅ − 1 . The rule stating that a b = a b only holds when both a and b are positive numbers.

By using the quadratic formula, you can even solve quadratic equations involving complex coefficients. In those cases, you need some knowledge about complex roots in order to simplify the expression inside the square root.

Solve 1 2 z 2 + i z + 3 2 i = 0 for z

You recognize the coefficients as a = 1 2 , b = i and c = 3 2 i . This means that you can use the quadratic formula to find the solutions:

All complex numbers w have two square roots. In this case you only need to consider the root whose argument lies in the interval [ 0 , π ) . The reason for this is that both solutions are included in ± w .

In order to find the square root of w = − 1 − 3 i , you first need to write w in polar form . In this case, the norm of w is r = 2 , while the argument is 𝜃 = 4 π 3 . Thus w = 2 e i 4 π 3 in polar form. You can now find the square root of w by taking the square root of the norm of w and dividing the argument of w by 2 . Thus the square root of w is 2 e i 2 π 3 . In Cartesian form, the square root is written as − 2 2 + 6 2 i . Thus the solutions to the equation are:

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1.5: Quadratic Equations with Complex Roots

• Last updated
• Save as PDF
• Page ID 40893

• Richard W. Beveridge
• Clatsop Community College

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In Section \(1.3,\) we considered the solution of quadratic equations that had two real-valued roots. This was due to the fact that in calculating the roots for each equation, the portion of the quadratic formula that is square rooted (\(b^{2}-4 a c,\) often called the discriminant) was always a positive number.

For example, in using the quadratic formula to calculate the the roots of the equation \(x^{2}-6 x+3=0,\) the discriminant is positive and we will end up with two real-valued roots: \[ \begin{array}{c} x^{2}-6 x+3=0 \\ a=1, b=-6, c=3 \\ =\frac{-(-6) \pm \sqrt{(-6)^{2}-4(1)(3)}}{2 * 1} \\ =\frac{6 \pm \sqrt{36-12}}{2} \\ =\frac{6 \pm \sqrt{24}}{2} \\ =\frac{6 \pm 4.899}{2} \end{array} \] \begin{array}{ll} \approx \frac{6+4.899}{2} & \approx \frac{6-4.899}{2} \\ \approx \frac{10.899}{2} & \approx \frac{1.101}{2} \\ \approx 5.449 & \approx 0.551 \end{array}

When we added and subtracted the square root of 24 to 6 in the quadratic formula, this created two answers, and they were real-valued because the square root of 24 is real-valued.

Another way to see this is graphically. If we graph \(y=x^{2}-6 x+3\) and find the \(x\) values that make \(y=0,\) these will appear along the \(x\) -axis, and will be the same values that solve the equation \(x^{2}-6 x+3=0\)

If we consider a related, but slightly different equation to start with, these relationships between the roots, the discriminant and the graphical intersections will be slightly different. \[ \begin{array}{c} x^{2}-6 x+9=0 \\ a=1, b=-6, c=9 \end{array} \]

\begin{aligned} x &=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(1)(9)}}{2 * 1} \\ &=\frac{6 \pm \sqrt{36-36}}{2} \\ &=\frac{6 \pm \sqrt{0}}{2} \\ &=\frac{6}{2}=3 \end{aligned}

Because the discriminant was 0 in this problem, we only get one real-valued answer.

Graphically, the additional 6 that was added to the original equation to change it from \(x^{2}-6 x+3\) to \(x^{2}-6 x+9\) shifts every \(y\) value on the graph up 6 units.

If we add an additional three units to the constant term of this quadratic equation, we encounter a third possibility.

\begin{array}{c} x^{2}-6 x+12=0 \\ a=1, b=-6, c=12 \\ =\frac{-(-6) \pm \sqrt{(-6)^{2}-4(1)(12)}}{2 * 1} \\ =\frac{6 \pm \sqrt{36-48}}{2} \\ =\frac{6 \pm \sqrt{-12}}{2} \\ =\frac{6}{2} \pm \frac{6 \cdot 464 i}{2} \\ \approx 3 \pm 1.732 i \end{array}

Here the discriminant is negative, which leads to two complex-valued answers. If the equation has real-valued coefficients, the complex roots will always come in conjugate pairs. Complex conjugates share the same real-valued part and have opposite signs in their complex-valued (or imaginary) parts: \(a \pm b i\)

Graphically, the previous problem was one step away from not intersecting the \(x\) -axis at all and the additional three units that we added on to get \(y=x^{2}-\) \(6 x+12\) moves the graph entirely away from the \(x\) -axis. Because the roots are complex-valued, we don't see any roots on the \(x\) -axis. The \(x\) -axis contains only real numbers.

since the calculator has been programmed for the quadratic formula, the focus of the problems in this section will be on putting them into standard form.

Example \(\PageIndex{1}\)

Solve for \(x\) \((2 x+1)(x+5)-2 x(x+7)=5(x+3)^{2}\)

\[ \begin{array}{c} (2 x+1)(x+5)-2 x(x+7)=5(x+3)^{2} \\ 2 x^{2}+11 x+5-2 x^{2}-14 x=5(x+3)(x+3) \\ -3 x+5=5\left(x^{2}+6 x+9\right) \\ -3 x+5=5 x^{2}+30 x+45 \\ 0=5 x^{2}+33 x+40 \\ x=5, b=33, c=40 \\ x=-5,-1.6 \end{array} \]

The fact that the roots of this equation were rational numbers means that the equation could have been solved by factoring. \[ \begin{array}{cc} 0=5 x^{2}+33 x+40 \\ 0=(5 x+8)(x+5) \\ 5 x=-8 & x+5=0 \\ 5 x+8=0 & x=-5 \\ x=-1.6 & \end{array} \]

Solve for \(x\) \((x-2)^{2}+3(4 x-1)(x+1) &=7(x+1)(x-1)\)

\[ \begin{aligned} x^{2}-4 x+4+3\left(4 x^{2}+3 x-1\right) &=7\left(x^{2}-1\right) \\ x^{2}-4 x+4+12 x^{2}+9 x-3 &=7 x^{2}-7 \\ 13 x^{2}+5 x+1 &=7 x^{2}-7 \\ 6 x^{2}+5 x+8 &=0 \\ a=6, b=5, c=8 & \\ x \approx-0.41 \overline{6} \pm 1.077 i \approx-\frac{5}{12} \pm 1.077 i \end{aligned} \]

Exercise \(\PageIndex{1}\)

Solve for \(x\) in each equation. Round any irrational values to the nearest 1000 th. 1) \(\quad 3 x^{2}-3 x=4\) 2) \(\quad 4 x^{2}-2 x=7\) 3) \(\quad 5 x^{2}=3-7 x\) 4) \(\quad 3 x^{2}=21-14 x\) 5) \(\quad 6 x^{2}+1=2 x\) 6) \(\quad 5 x-3 x^{2}=17\) 7) \(\quad (5 x-1)(2 x+3)=3 x-20\) 8) \(\quad (x+4)(3 x-1)=9 x-5\) 9) \(\quad (x-2)^{2}=8 x(x-1)+10\) 10) \(\quad (2 x-3)^{2}=2 x-7 x^{2}\) 11) \(\quad (x+5)(x-6)=(2 x-1)(x-4)\) 12) \(\quad (3 x-4)(x+2)=(2 x-5)(x+5)\)

1) \(\quad x \approx 1.758,-0.758\) 3) \(\quad x \approx 0.344,-1.744\) 5) \(\quad x \approx 0.1 \overline{6} \pm 0.373 i\) 7) \(\quad x \approx-0.5 \pm 1.204 i\) 9) \(\quad x \approx 0.286 \pm 0.881 i\) 11) \(\quad x \approx 4 \pm 4.243 i\)

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