Question

question 10 of 10, step 1 of 1 solve the following radical equation. (sqrt{7x + 11}=sqrt{4x + 1}+2) answer how to enter your answer (opens in new window) write your answer(s) beginning with the first answer box. if applicable, the second answer box may be left blank. (x=)

Explanation:

Step1: Isolate one radical term

$\sqrt{7x+11} - 2 = \sqrt{4x+1}$

Step2: Square both sides

$(\sqrt{7x+11} - 2)^2 = (\sqrt{4x+1})^2$
$7x+11 - 4\sqrt{7x+11} + 4 = 4x+1$

Step3: Simplify the equation

$7x + 15 - 4\sqrt{7x+11} = 4x + 1$
$3x + 14 = 4\sqrt{7x+11}$

Step4: Square both sides again

$(3x + 14)^2 = (4\sqrt{7x+11})^2$
$9x^2 + 84x + 196 = 16(7x+11)$
$9x^2 + 84x + 196 = 112x + 176$

Step5: Rearrange to quadratic form

$9x^2 - 28x + 20 = 0$

Step6: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=9$, $b=-28$, $c=20$:
$\Delta = (-28)^2 - 4\times9\times20 = 784 - 720 = 64$
$x=\frac{28\pm\sqrt{64}}{18}=\frac{28\pm8}{18}$
$x_1=\frac{28+8}{18}=\frac{36}{18}=2$, $x_2=\frac{28-8}{18}=\frac{20}{18}=\frac{10}{9}$

Step7: Verify solutions

For $x=2$:
Left: $\sqrt{7\times2+11}=\sqrt{25}=5$; Right: $\sqrt{4\times2+1}+2=\sqrt{9}+2=3+2=5$. Valid.
For $x=\frac{10}{9}$:
Left: $\sqrt{7\times\frac{10}{9}+11}=\sqrt{\frac{70}{9}+\frac{99}{9}}=\sqrt{\frac{169}{9}}=\frac{13}{3}$;
Right: $\sqrt{4\times\frac{10}{9}+1}+2=\sqrt{\frac{40}{9}+\frac{9}{9}}+2=\sqrt{\frac{49}{9}}+2=\frac{7}{3}+2=\frac{13}{3}$. Valid.

Answer:

$x = 2$, $\frac{10}{9}$