Back to Exercise: Solve quadratic equations with complex solutions

Exercises: Solve Quadratic Equations with Complex Solutions

Show all steps. Write complex solutions in $a + bi$ form. Include both solutions.

Grade 9·20 problems·~40 min·Common Core Math - HS Number and Quantity·standard·hsn-cn-c-7

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Warm-Up: Review What You Know

For the quadratic $ax^2 + bx + c = 0$ , the discriminant is $b^2 - 4ac$ . If the discriminant is negative, which statement is correct?

The equation has two distinct real solutions.

The equation has no solutions at all.

The equation has two complex (non-real) solutions.

The equation has one repeated real solution.

The quadratic formula is $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

For $x^2 - 6x + 9 = 0$ , what is the discriminant $b^2 - 4ac$ ?

$0$

$-9$

$45$

$72$

Simplify $\sqrt{-9}$ . Write it in the form $ki$ where $k$ is a positive real number. Enter the value of $k$ .

Fluency Practice

Compute the discriminant $b^2 - 4ac$ for $x^2 + 4x + 13 = 0$ . Which case applies?

Discriminant $= -36 < 0$ — two complex conjugate solutions

Discriminant $= 32 > 0$ — two distinct real solutions

Discriminant $= 0$ — one repeated real solution

Discriminant $= -48 < 0$ — no solutions

Solve $x^2 - 4x + 5 = 0$ using the quadratic formula. The solutions are $x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\pm$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $i$ .

Solve $x^2 + 2x + 5 = 0$ . The solutions are $x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\pm$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $i$ .

Solve $2x^2 - 2x + 5 = 0$ . The solutions are $x = $ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\pm$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $i$ .

A quadratic equation with real coefficients has $3 + 4i$ as one solution. What must the other solution be?

$3 + 4i$ (the same solution again)

$-3 - 4i$ (the negative)

$3 - 4i$ (the conjugate)

$4 + 3i$ (parts swapped)

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