Question

question 1 of 10
which of the following are solutions to the equation below?
check all that apply.
$9x^2 - 6x + 1 = 7$

a. $x = -\frac{2sqrt{2}}{3}$
b. $x = \frac{sqrt{7}}{3} + 1$
c. $x = \frac{-sqrt{7} + 1}{3}$
d. $x = \frac{sqrt{7} + 1}{3}$
e. $x = \frac{2sqrt{2}}{3}$
f. $x = \frac{-sqrt{7}}{3} + 1$

Explanation:

Step1: Rearrange the equation

First, we start with the equation \(9x^{2}-6x + 1=7\). Subtract 7 from both sides to get it in standard quadratic form \(ax^{2}+bx + c = 0\).
\(9x^{2}-6x+1 - 7=0\)
\(9x^{2}-6x - 6 = 0\)
We can simplify this equation by dividing all terms by 3:
\(3x^{2}-2x - 2=0\)

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
In our equation \(3x^{2}-2x - 2 = 0\), we have \(a = 3\), \(b=- 2\), and \(c=-2\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\).
\(\Delta=(-2)^{2}-4\times3\times(-2)\)
\(=4 + 24\)
\(=28\)
Then, find the square root of the discriminant: \(\sqrt{\Delta}=\sqrt{28} = 2\sqrt{7}\) (Wait, no, wait. Wait, in our original equation after simplifying we had \(3x^{2}-2x - 2=0\), but wait, let's go back. Wait, the original equation was \(9x^{2}-6x + 1=7\), which is a perfect square trinomial on the left. \(9x^{2}-6x + 1=(3x - 1)^{2}\). So let's redo that step.

Step1 (corrected): Recognize the perfect square

The left - hand side of the equation \(9x^{2}-6x + 1\) is a perfect square trinomial. Recall that \((a - b)^{2}=a^{2}-2ab + b^{2}\). For \(9x^{2}-6x + 1\), \(a = 3x\) and \(b = 1\) since \((3x)^{2}=9x^{2}\), \(2\times3x\times1 = 6x\) and \(1^{2}=1\). So \(9x^{2}-6x + 1=(3x - 1)^{2}\).
So our equation becomes \((3x - 1)^{2}=7\)

Step2 (corrected): Take square roots

Take the square root of both sides: \(3x-1=\pm\sqrt{7}\)

Step3 (corrected): Solve for x

Case 1: \(3x-1=\sqrt{7}\)
Add 1 to both sides: \(3x=\sqrt{7}+1\)
Divide both sides by 3: \(x=\frac{\sqrt{7}+1}{3}\)

Case 2: \(3x - 1=-\sqrt{7}\)
Add 1 to both sides: \(3x=-\sqrt{7}+1\)
Divide both sides by 3: \(x=\frac{-\sqrt{7}+1}{3}\)

Answer:

C. \(x=\frac{-\sqrt{7}+1}{3}\), D. \(x=\frac{\sqrt{7}+1}{3}\)