Question

1. what are the solutions to

$x^2 + 2x - 24 = 0$
a. $x = 6$ and $x = -4$
b. $x = -6$ and $x = 4$
c. $x = -6$ and $x = -4$
d. $x = 6$ and $x = 4$

1. what are the solutions to

$5x^2 - 180 = 0$
a. $x = 9$ and $x = -9$
b. $x = 18$ and $x = -18$
c. $x = 6$ and $x = -6$
d. $x = 4$ and $x = -4$

Explanation:

Problem 1:

Step1: Factor the quadratic equation

We have the equation \(x^2 + 2x - 24 = 0\). We need to find two numbers that multiply to \(-24\) and add up to \(2\). The numbers are \(6\) and \(-4\) since \(6\times(-4)=-24\) and \(6 + (-4)=2\). So we can factor the equation as \((x + 6)(x - 4)=0\).

Step2: Solve for \(x\)

Using the zero - product property, if \((x + 6)(x - 4)=0\), then either \(x+6 = 0\) or \(x - 4=0\).

• If \(x+6 = 0\), then \(x=-6\).
• If \(x - 4=0\), then \(x = 4\).

Step1: Simplify the quadratic equation

We have the equation \(5x^2-180 = 0\). First, we can divide the entire equation by \(5\) to simplify it. \(\frac{5x^2}{5}-\frac{180}{5}=\frac{0}{5}\), which gives us \(x^2-36 = 0\).

Step2: Factor the equation

The equation \(x^2 - 36=0\) is a difference of squares, and we can factor it as \((x + 6)(x - 6)=0\) (since \(a^2-b^2=(a + b)(a - b)\) and here \(a=x\), \(b = 6\) as \(x^2=x\times x\) and \(36 = 6\times6\)).

Step3: Solve for \(x\)

Using the zero - product property, if \((x + 6)(x - 6)=0\), then either \(x+6 = 0\) or \(x - 6=0\).

• If \(x+6 = 0\), then \(x=-6\).
• If \(x - 6=0\), then \(x = 6\).

Answer:

b. \(x=-6\) and \(x = 4\)

Problem 2: