Question

cubic and cube root functions and equations
what is the solution to the equation $sqrt{4t + 5} = 3 - sqrt{t + 5}$?
$x = -1, x = 11$
$x = -1$
$x = 11$
no solution

Explanation:

Step1: Let $u=\sqrt{t+5}$

$u \geq 0$, and $\sqrt{4t+5}=\sqrt{4(u^2-5)+5}=\sqrt{4u^2-15}$

Step2: Rewrite the equation

$\sqrt{4u^2-15}=3-u$

Step3: Square both sides

$4u^2-15=(3-u)^2$
$4u^2-15=9-6u+u^2$

Step4: Simplify to quadratic

$3u^2+6u-24=0$
$u^2+2u-8=0$

Step5: Solve quadratic equation

$(u+4)(u-2)=0$
$u=-4$ or $u=2$

Step6: Check valid $u$

$u=-4$ is invalid ($u\geq0$). For $u=2$:
Right-hand side: $3-2=1$
Left-hand side: $\sqrt{4(2)^2-15}=\sqrt{16-15}=1$
Now substitute back $u=\sqrt{t+5}=2$:
$t+5=4 \implies t=-1$

Step7: Verify $t=-1$ in original equation

Left: $\sqrt{4(-1)+5}=\sqrt{1}=1$
Right: $3-\sqrt{-1+5}=3-2=1$
Wait, correction: $t=-1$ is valid. So the answer is $x=-1$ (using $x$ as in options)

Step1: Let $u=\sqrt{t+5}$

$u\geq0$, $\sqrt{4t+5}=\sqrt{4u^2-15}$

Step2: Rewrite original equation

$\sqrt{4u^2-15}=3-u$

Step3: Square both sides

$4u^2-15=9-6u+u^2$

Step4: Simplify to quadratic

$3u^2+6u-24=0 \implies u^2+2u-8=0$

Step5: Solve for $u$

$(u+4)(u-2)=0 \implies u=2$ (valid, $u\geq0$)

Step6: Find $t$ from $u=2$

$\sqrt{t+5}=2 \implies t+5=4 \implies t=-1$

Step7: Verify solution

Left: $\sqrt{4(-1)+5}=1$, Right: $3-\sqrt{-1+5}=1$. Both equal.

Answer:

$x = 1$ (Note: The variable in the question is $t$, so correct answer is $t=1$; among the given options, the closest valid solution after checking is $x=11$ is invalid, $x=-1$ is invalid, so no solution is incorrect, correction: the correct solution is $t=1$, but since it's not listed, rechecking shows:
Wait, re-solving correctly: