$x^2 + 12x - 36 = 0$
To find the roots of the quadratic equation \(x^2 + 12x - 36 = 0\), we can use the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Here, the coefficients of the equation \(ax^2 + bx + c = 0\) are:
- \(a = 1\)
- \(b = 12\)
- \(c = -36\)
Substitute these values into the quadratic formula:
\(x = \frac{-12 \pm \sqrt{(12)^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1}\)
\(x = \frac{-12 \pm \sqrt{144 + 144}}{2}\)
\(x = \frac{-12 \pm \sqrt{288}}{2}\)
Simplify \(\sqrt{288}\):
\(\sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2}\)
Now, substitute back:
\(x = \frac{-12 \pm 12\sqrt{2}}{2}\)
Divide each term by 2:
\(x = \frac{-12}{2} \pm \frac{12\sqrt{2}}{2}\)
\(x = -6 \pm 6\sqrt{2}\)
Therefore, the roots are:
- \(x_1 = -6 + 6\sqrt{2}\)
- \(x_2 = -6 - 6\sqrt{2}\)
Upon inspection, we can factor \(x_1\) and \(x_2\) with \(6(\sqrt{2} - 1)\):
- \(x_1 = -6 + 6\sqrt{2} = 6(\sqrt{2} - 1)\)
- \(x_2 = -6 - 6\sqrt{2} = -6(1 + \sqrt{2})\)
Thus, the roots of the equation \(x^2 + 12x - 36 = 0\) are
$-6(1 + \sqrt{2}), 6(\sqrt{2} - 1)$
, which matches the correct answer. Hence, the option
$-6(1 + \sqrt{2}), 6(\sqrt{2} - 1)$ is correct.
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