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All courses Algebra II · N-CN.7 5 of 55
Solve real-coefficient quadratic equations that have complex solutions.

Classify any quadratic using the discriminant

Problem
For \(x^{2}+4x+8=0\), calculate the discriminant \(b^{2}-4ac\) and use its sign to classify the number and type of solutions over the complex numbers.
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Hint

Copy a,b,c with signs before evaluating b²−4ac.

Δ>0: two distinct real; Δ=0: one repeated real; Δ<0: two nonreal conjugates for real coefficients.

Solution walkthrough

01

Source the signed coefficients

\[x^2+4x+8=0~->~a=1;~b=4;~c=8\]

The discriminant applies because the equation is a quadratic in standard form ax squared plus bx plus c equals zero.

02

Substitute into the discriminant

\[D=b^2-4ac=4^2-4(1)(8)\]

Each value comes directly from the corresponding coefficient, including the leading coefficient 1.

03

Calculate and interpret the sign

\[D=16-32=-16<0\]

A negative discriminant makes the square-root term nonreal; with real coefficients, the two roots form a conjugate pair.

04

Classify and verify

\[D<0~->~\text{two}~\text{distinct}~\text{nonreal}~\text{complex}~\text{conjugate}~\text{solutions}\]

Indeed, the quadratic formula gives x=(-4 plus or minus 4i)/2=-2 plus or minus 2i, confirming two distinct conjugates.

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Another way

  1. Complete the square: (x+2)²=-4, so x=-2 plus or minus 2i; this confirms the discriminant classification.

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Common mistake

Do not report no solutions. A negative discriminant means no real solutions, but over the complex numbers it gives two nonreal conjugate solutions.

Solution walkthrough video