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

1 answer

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

\[ \boxed{1} \]