Resolved: "Thesolution set of the equation (3x−4)2−2=11"

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

amanf

Answered question

2021-10-02

Thesolution set of the equation

{(3 x - 4)}^{2} - 2 = 11

Answer & Explanation

2abehn

Skilled2021-10-03Added 88 answers

Step 1
Consider the given equation,
${(3 x - 4)}^{2} - 2 = 11$
Add 2 to both sides:
${(3 x - 4)}^{2} - 2 = 11$
${(3 x - 4)}^{2} - 2 + 2 = 11 + 2$
${(3 x - 4)}^{2} = 13$
Now, apply the square root propertyin above equation.
${(3 x - 4)}^{2} = 13$
$(3 x - 4) = \pm \sqrt{13}$
Now, add 4 to the both side of the above equation,
$(3 x - 4) = \pm \sqrt{13}$
$3 x - 4 + 4 = \pm \sqrt{13} + 4$
$3 x = 4 \pm \sqrt{13}$
$x = \frac{4 \pm \sqrt{13}}{3}$
Check:
Substitute the value of $x = \frac{4 + \sqrt{13}}{3}$ in the provided equation ${(3 x - 4)}^{2} - 2 = 11$
${(3 (\frac{4 + \sqrt{13}}{3}) - 4)}^{2} - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$(4 + \sqrt{13} - 4)^{2} - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$13 - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$11 \begin{matrix} ? \\ = \end{matrix} 11$
which is true.
Now, substitute the value of $x = \frac{4 - \sqrt{13}}{3}$ in the provided equation ${(3 x - 4)}^{2} - 2 = 11$
${(3 (\frac{4 + \sqrt{13}}{3}) - 4)}^{2} - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$(4 - \sqrt{13} - 4)^{2} - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$13 - 2 \begin{matrix} ? \\ = \end{matrix} 11$
$11 \begin{matrix} ? \\ = \end{matrix} 11$
which is true.
Hence, the solution set of the equation ${(3 x - 4)}^{2} - 2 = 11$ is ${\frac{4 \pm \sqrt{13}}{3}}$

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