Question

try: find the zeros of the quadratic $x^2 + 22x = -121$

try: find the roots of the quadratic $x^2 - 12x - 2 = 0$

Explanation:

First Problem: Find the zeros of \( x^2 + 22x = -121 \)

Step1: Rearrange to standard form

Move all terms to left: \( x^2 + 22x + 121 = 0 \)

Step2: Factor the quadratic

Notice it's a perfect square: \( (x + 11)^2 = 0 \)

Step3: Solve for \( x \)

Take square root: \( x + 11 = 0 \Rightarrow x = -11 \) (double root)

Step1: Identify \( a, b, c \)

For \( ax^2 + bx + c = 0 \), \( a = 1 \), \( b = -12 \), \( c = -2 \)

Step2: Use quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Calculate discriminant: \( \Delta = (-12)^2 - 4(1)(-2) = 144 + 8 = 152 \)
Simplify \( \sqrt{152} = \sqrt{4 \times 38} = 2\sqrt{38} \)

Step3: Substitute into formula

\( x = \frac{12 \pm 2\sqrt{38}}{2} = 6 \pm \sqrt{38} \)

Answer:

The zero of the quadratic is \( x = -11 \) (with multiplicity 2)

Second Problem: Find the roots of \( x^2 - 12x - 2 = 0 \)