QuestionSolve the equation $x^{2}+4x-4=0$ .

Studdy Solution

STEP 1

Question: What is the form of a quadratic equation?
A) $ax^3 + bx + c = 0$ B) $ax^2 + bx + c = 0$ C) $ax + b = 0$ D) $a/bx + c = 0$
Answer: B) $ax^2 + bx + c = 0$

STEP 2

Question: In the equation $x^{2}+4x-4=0$ , what are the coefficients $a$ , $b$ , and $c$ ?
A) $a=2$ , $b=4$ , $c=-4$ B) $a=1$ , $b=2$ , $c=-4$ C) $a=1$ , $b=4$ , $c=-4$ D) $a=4$ , $b=1$ , $c=-4$
Answer: C) $a=1$ , $b=4$ , $c=-4$

STEP 3

Question: Substitute the coefficients into the quadratic formula. What do you get?
A) $x = \frac{-4 \pm \sqrt{4^{2}-4(2)(-4)}}{2(1)}$ B) $x = \frac{-4 \pm \sqrt{4^{2}-4(1)(4)}}{2(1)}$ C) $x = \frac{-4 \pm \sqrt{4^{2}-4(1)(-4)}}{2(1)}$ D) $x = \frac{-1 \pm \sqrt{4^{2}-4(4)(-4)}}{2(4)}$
Answer: C) $x = \frac{-4 \pm \sqrt{4^{2}-4(1)(-4)}}{2(1)}$

STEP 4

Question: Simplify the expression under the square root in the quadratic formula. What is the simplified expression?
A) $x = \frac{-4 \pm \sqrt{16-16}}{2}$ B) $x = \frac{-4 \pm \sqrt{16+16}}{2}$ C) $x = \frac{-4 \pm \sqrt{16-4}}{2}$ D) $x = \frac{-4 \pm \sqrt{16+4}}{2}$
Answer: B) $x = \frac{-4 \pm \sqrt{16+16}}{2}$

STEP 5

Question: Simplify the final expression to find the roots of the quadratic equation. What are the roots?
A) $x = -2 \pm 4\sqrt{2}$ B) $x = -2 \pm \sqrt{2}$ C) $x = -2 \pm 2\sqrt{2}$ D) $x = -4 \pm 2\sqrt{2}$
Answer: C) $x = -2 \pm 2\sqrt{2}$

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