Question

solve the quadratic equation.
$4x^2 - 9 = 0$
a. $\bigcirc \frac{-3}{2}$
b. $\bigcirc \frac{3}{2}$
c. $\bigcirc \frac{3}{3}$ and $\frac{-3}{3}$
d. $\bigcirc \frac{9}{4}$ and $\frac{-9}{4}$
e. $\bigcirc \frac{3}{2}$ and $\frac{-3}{2}$

Explanation:

Step1: Isolate the squared term

Given the equation \(4x^2 - 9 = 0\), we first add 9 to both sides to isolate the \(x^2\) term. So we get \(4x^2=9\).

Step2: Solve for \(x^2\)

Then, divide both sides by 4 to solve for \(x^2\). This gives us \(x^2=\frac{9}{4}\).

Step3: Take square roots

To solve for \(x\), we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution. So \(x = \pm\sqrt{\frac{9}{4}}\).

Step4: Simplify the square root

Simplifying \(\sqrt{\frac{9}{4}}\), we know that \(\sqrt{9} = 3\) and \(\sqrt{4}=2\), so \(\sqrt{\frac{9}{4}}=\frac{3}{2}\). Therefore, \(x=\frac{3}{2}\) or \(x = -\frac{3}{2}\).

Answer:

E. \(\frac{3}{2}\) and \(-\frac{3}{2}\)