Question

cc.2.2.hs.d.10
solve for x:
$2x^2 - 8x = 1$

$2 \pm (3/2)\sqrt{2}$
$2 \pm 3\sqrt{2}$
$-2 \pm (2/3)\sqrt{2}$
$-2 \pm 3\sqrt{2}$

Explanation:

Step1: Divide by 2 to simplify

Divide the entire equation \(2x^{2}-8x = 1\) by 2. We get \(x^{2}-4x=\frac{1}{2}\).

Step2: Complete the square

For the quadratic expression \(x^{2}-4x\), to complete the square, we take half of the coefficient of \(x\) (which is \(- 4\)), square it. Half of \(-4\) is \(-2\), and squaring it gives \((-2)^{2}=4\). Add and subtract 4 on the left - hand side:
\(x^{2}-4x + 4-4=\frac{1}{2}\)
Rewrite \(x^{2}-4x + 4\) as \((x - 2)^{2}\), so the equation becomes \((x - 2)^{2}-4=\frac{1}{2}\).

Step3: Isolate the square term

Add 4 to both sides of the equation: \((x - 2)^{2}=\frac{1}{2}+4\).
Calculate \(\frac{1}{2}+4=\frac{1 + 8}{2}=\frac{9}{2}\). So \((x - 2)^{2}=\frac{9}{2}\).

Step4: Take square roots

Take the square root of both sides: \(x - 2=\pm\sqrt{\frac{9}{2}}\).
Simplify \(\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\) (rationalizing the denominator).

Step5: Solve for x

Add 2 to both sides of the equation: \(x=2\pm\frac{3\sqrt{2}}{2}=2\pm\frac{3}{2}\sqrt{2}\).

Answer:

\(2\pm\frac{3}{2}\sqrt{2}\) (corresponding to the option "2 ± (3/2)√2")