Question

find the zeros of the quadratic equation:
$x^2 + 7x + 12 = 0$
$\circ -3, 4$
$\circ 3, -4$
$\circ 3, 4$
$\circ -3, -4$

question 24
2.5 pts
a student is solving the quadratic equation below by completing the square. which of the following equations shows an accurate step in the process?
$x^2 + 12x + 18 = 0$
$\circ (x + 9)^2 = 18$
$\circ (x + 6)^2 = 54$
$\circ (x + 9)^2 = 54$
$\circ (x + 6)^2 = 18$

Explanation:

First Problem: Find the zeros of $x^2+7x+12=0$

Step1: Factor the quadratic

We find two numbers that multiply to 12 and add to 7: 3 and 4.
$x^2+7x+12=(x+3)(x+4)=0$

Step2: Solve for $x$

Set each factor equal to 0.
$x+3=0 \implies x=-3$; $x+4=0 \implies x=-4$

Step1: Isolate constant term

Subtract 18 from both sides.
$x^2+12x=-18$

Step2: Complete the square

Take half of 12 (which is 6), square it ($6^2=36$), add to both sides.
$x^2+12x+36=-18+36$

Step3: Rewrite as squared binomial

Left side is a perfect square trinomial.
$(x+6)^2=18$

Answer:

-3, -4

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Second Problem: Completing the square for $x^2+12x+18=0$