what is the solution to the equation $sqrt{4t + 5}=3-sqrt{t + 5}$?\n$x=-1,x = 11$\n$x=-1$\n$x = 11$\nno…

Explanation:

Step1: Isolate one square - root term

Isolate $\sqrt{4t + 5}$ on the left - hand side. The given equation is $\sqrt{4t+5}=3-\sqrt{t + 5}$.

Step2: Square both sides

$(\sqrt{4t + 5})^2=(3-\sqrt{t + 5})^2$. Using the formula $(a - b)^2=a^2-2ab + b^2$, we get $4t+5 = 9-6\sqrt{t + 5}+(t + 5)$.

Step3: Simplify the equation

$4t+5=9-6\sqrt{t + 5}+t + 5$. Combine like terms: $4t+5=t + 14-6\sqrt{t + 5}$. Rearrange to isolate the square - root term: $6\sqrt{t + 5}=t + 14-(4t + 5)$. $6\sqrt{t + 5}=t + 14-4t - 5$. $6\sqrt{t + 5}=-3t + 9$.

Step4: Square both sides again

$(6\sqrt{t + 5})^2=(-3t + 9)^2$. $36(t + 5)=9t^2-54t + 81$. Expand the left - hand side: $36t+180=9t^2-54t + 81$. Rearrange to form a quadratic equation: $9t^2-54t-36t + 81 - 180 = 0$. $9t^2-90t-99 = 0$. Divide through by 9: $t^2-10t - 11 = 0$.

Step5: Solve the quadratic equation

Factor the quadratic equation: $(t - 11)(t+1)=0$. Set each factor equal to zero: $t - 11 = 0$ gives $t = 11$; $t + 1 = 0$ gives $t=-1$.

Step6: Check for extraneous solutions

When $t = 11$: Left - hand side of the original equation $\sqrt{4\times11+5}=\sqrt{44 + 5}=\sqrt{49}=7$. Right - hand side: $3-\sqrt{11 + 5}=3-\sqrt{16}=3 - 4=-1$. Since $7\neq-1$, $t = 11$ is an extraneous solution. When $t=-1$: Left - hand side $\sqrt{4\times(-1)+5}=\sqrt{-4 + 5}=1$. Right - hand side $3-\sqrt{-1 + 5}=3-\sqrt{4}=3 - 2 = 1$. So $t=-1$ is a valid solution.

Answer:

$x=-1$