Question

question 8
solve by using the quadratic formula.
$x^2 + 6x - 36 = 0$
$3 \pm 3\sqrt{5}$
$3 \pm 3\sqrt{10}$
$-6 \pm 3\sqrt{10}$
$-3 \pm 3\sqrt{5}$

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For the equation \(x^2+6x - 36 = 0\), we have \(a = 1\), \(b = 6\), and \(c=-36\).

Step2: Calculate Discriminant

First, calculate the discriminant \(D=b^2 - 4ac\). Substitute \(a = 1\), \(b = 6\), \(c=-36\) into the formula:
\(D=(6)^2-4\times1\times(-36)=36 + 144 = 180\)

Step3: Apply Quadratic Formula

Now, substitute \(a = 1\), \(b = 6\), and \(D = 180\) into the quadratic formula:
\(x=\frac{-6\pm\sqrt{180}}{2\times1}=\frac{-6\pm\sqrt{36\times5}}{2}=\frac{-6\pm6\sqrt{5}}{2}\)
Simplify the fraction:
\(x=-3\pm3\sqrt{5}\) (Wait, there is a mistake above. Wait, \(\sqrt{180}=\sqrt{36\times5}=6\sqrt{5}\)? No, wait \(180 = 36\times5\)? No, \(36\times5 = 180\), but wait the options have \(\sqrt{5}\) or \(\sqrt{10}\)? Wait, let's recalculate discriminant. Wait, the equation is \(x^2+6x - 36=0\), so \(b^2-4ac=6^2-4\times1\times(-36)=36 + 144 = 180\). Wait, but \(180=36\times5\)? No, \(36\times5 = 180\), but \(180 = 36\times5\), but \(\sqrt{180}=\sqrt{36\times5}=6\sqrt{5}\), then \(x=\frac{-6\pm6\sqrt{5}}{2}=-3\pm3\sqrt{5}\). Wait, but the last option is \(-3\pm3\sqrt{5}\). Wait, maybe I made a mistake in discriminant. Wait, no, let's check again. The equation is \(x^2 + 6x - 36 = 0\). So \(a = 1\), \(b = 6\), \(c=-36\). Then \(b^2-4ac=36-4(1)(-36)=36 + 144 = 180\). \(\sqrt{180}=\sqrt{36\times5}=6\sqrt{5}\). Then dividing by 2: \(\frac{-6\pm6\sqrt{5}}{2}=-3\pm3\sqrt{5}\). So the correct option is the last one: \(-3\pm3\sqrt{5}\).

Answer:

\(\boldsymbol{-3\pm3\sqrt{5}}\) (Corresponding to the last option: \(-3\pm3\sqrt{5}\))