Question

Solve 7-square root of 4x+5=x+3

Answer 1

To solve the equation \( 7 - \sqrt{4x + 5} = x + 3 \), we first isolate the square root term:

\[
7 - x - 3 = \sqrt{4x + 5}
\]

This simplifies to:

\[
4 - x = \sqrt{4x + 5}
\]

Next, we square both sides of the equation to eliminate the square root:

\[
(4 - x)^2 = 4x + 5
\]

Expanding the left side:

\[
16 - 8x + x^2 = 4x + 5
\]

Rearranging all terms to one side of the equation gives:

\[
x^2 - 12x + 11 = 0
\]

Now, we can factor the quadratic:

\[
(x - 11)(x - 1) = 0
\]

Setting each factor to zero gives the potential solutions:

\[
x - 11 = 0 \quad \text{or} \quad x - 1 = 0
\]

This means:

\[
x = 11 \quad \text{or} \quad x = 1
\]

Next, we must check both solutions in the original equation to ensure they are valid:

**For \( x = 11 \):**

\[
7 - \sqrt{4(11) + 5} = 11 + 3
\]
\[
7 - \sqrt{44 + 5} = 14
\]
\[
7 - \sqrt{49} = 14
\]
\[
7 - 7 = 14 \quad \text{(false)}
\]

**For \( x = 1 \):**

\[
7 - \sqrt{4(1) + 5} = 1 + 3
\]
\[
7 - \sqrt{4 + 5} = 4
\]
\[
7 - \sqrt{9} = 4
\]
\[
7 - 3 = 4 \quad \text{(true)}
\]

Thus, the only solution to the equation \( 7 - \sqrt{4x + 5} = x + 3 \) is

\[
\boxed{1}
\]