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Wize High School Algebra II Textbook (Common Core) > Quadratic Equations & Complex Numbers

Solving Quadratic Equations with Complex Solutions

Complex NumbersPrevious Section
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Quadratic Formula

When using the quadratic formula, you may have run into the issue of taking the square root of a negative number.

Wize Concept
Recall the quadratic formula for finding the roots of the real quadratic ax2+bx+c=0ax^2+bx+c=0:
x=−b±b2−4ac2a\boxed{\quad x=\dfrac{-b\pm\sqrt{\colorTwo{b^2-4ac}}}{2a} \quad}

b2−4ac\colorTwo{b^2-4ac} is called the discriminant. For any real quadratic:
  • b2−4ac≥0    ⟹  \colorTwo{b^2-4ac} \ge 0 \ \ \implies real solutions
  • b2−4ac<0    ⟹  \colorTwo{b^2-4ac} <0 \ \ \implies complex solutions (taking the square root of a negative number)
Wize Tip
Complex solutions always come in conjugate pairs!
If a+bia+bi is one root, then the other root is a−bia-bi.

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Example
Find the roots of the real quadratic p(x)=x2+2x+2p(x)=x^2+2x+2.
We want to find the values of xx such that x2+2x+2=0x^2+2x+2=0.
We have a=1,  b=2,  c=2a=1,\ \ b=2,\ \ c=2. Let's apply the quadratic formula.
x=−2±22−4(1)(2)2(1)=−2±−42=−2±−142=−2±i⋅22=−1±i\begin{aligned} x &= \dfrac{-2 \pm \sqrt{2^2-4(1)(2)}}{2(1)}\\[1em] &=\dfrac{-2 \pm \sqrt{-4}}{2}\\[1em] &=\dfrac{-2 \pm \colorOne{\sqrt{-1}}\sqrt{4}}{2}\\[1em] &=\dfrac{-2 \pm \colorOne{i}\cdot 2}{2}\\[1em] &= \boxed{-1 \pm i} \end{aligned}


Wize Concept
The quadratic formula works even if the quadratic has complex coefficients.
Note: you may need to find square roots of a complex number. Remember to use polar form to find the roots!

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Example: Roots of Real Quadratics

A real quadratic has a root at 2−3i2-3i . Find the other root along with the standard form of a quadratic with these roots.
Since we are given one root of a real quadratic and the root is complex, the other root is the complex conjugate: 2+3i\boxed{2+3i}.
We can find the real quadratic by multiplying the two factors that yield these roots:
(x−(2−3i))\Big(x- (2-3i) \Big) and (x−(2+3i))\Big(x- (2+3i) \Big) are the factors that, when set to be equal to zero, yield the desired roots.
  ⟹  p(x)=(x−(2−3i))(x−(2+3i))=x2−(2+3i)x−(2−3i)x+(2−3i)(2+3i)=x2−2x−3ix−2x+3ix+4+9=x2−4x+13\begin{aligned} \implies p(x) &= \Big(x- (2-3i) \Big)\Big(x- (2+3i) \Big)\\[0.5em] &= x^2 -(2+3i)x -(2-3i)x + (2-3i)(2+3i)\\[0.5em] &= x^2 -2x -\cancel{3ix} -2x+\cancel{3ix} + 4 + 9\\[0.5em] &= \boxed{x^2 -4x +13}\\[0.5em] \end{aligned}