Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it
solve the square root equation \\(\sqrt{7x - 12} = x\\). (1 point)
\\(\circ\\) \\(x = 2\\)
\\(\circ\\) \\(x = \frac{3}{2}\\)
\\(\circ\\) \\(x = 3\\) and \\(x = 4\\)
\\(\circ\\) \\(x = -4\\) and \\(x = -3\\)
check answer remaining attempts : 3

Explanation:

Step1: Square both sides

To eliminate the square root, square both sides of the equation \(\sqrt{7x - 12}=x\). We get \((\sqrt{7x - 12})^2=x^2\), which simplifies to \(7x - 12=x^2\).

Step2: Rearrange into quadratic form

Rearrange the equation \(7x - 12=x^2\) to standard quadratic form \(ax^2+bx + c = 0\). So, \(x^2-7x + 12 = 0\).

Step3: Factor the quadratic

Factor the quadratic equation \(x^2-7x + 12 = 0\). We need two numbers that multiply to 12 and add to -7. Those numbers are -3 and -4. So, \((x - 3)(x - 4)=0\).

Step4: Solve for x

Set each factor equal to zero: \(x - 3 = 0\) gives \(x = 3\), and \(x - 4 = 0\) gives \(x = 4\).

Step5: Check solutions

Check \(x = 3\): Substitute into the original equation \(\sqrt{7(3)-12}=\sqrt{21 - 12}=\sqrt{9}=3\), which is equal to \(x = 3\), so it's valid.
Check \(x = 4\): Substitute into the original equation \(\sqrt{7(4)-12}=\sqrt{28 - 12}=\sqrt{16}=4\), which is equal to \(x = 4\), so it's valid.

Answer:

\(x = 3\) and \(x = 4\) (corresponding to the option " \(x = 3\) and \(x = 4\)")