48. $sqrt{4x + 13}=2x - 1$

Explanation:

Step1: Square both sides

$(\sqrt{4x + 13})^2=(2x - 1)^2$ $4x+13 = 4x^2-4x + 1$

Step2: Rearrange to form a quadratic equation

$4x^2-4x-4x+1 - 13=0$ $4x^2-8x - 12 = 0$ Divide through by 4: $x^2-2x - 3=0$

Step3: Factor the quadratic equation

$(x - 3)(x+1)=0$

Step4: Solve for x

$x - 3=0$ gives $x = 3$; $x+1=0$ gives $x=-1$

Step5: Check for extraneous solutions

For $x = 3$: $\sqrt{4\times3+13}=\sqrt{12 + 13}=\sqrt{25}=5$, and $2\times3-1=6 - 1=5$, so $x = 3$ is a valid solution. For $x=-1$: $\sqrt{4\times(-1)+13}=\sqrt{-4 + 13}=\sqrt{9}=3$, and $2\times(-1)-1=-2 - 1=-3$. Since $3\neq - 3$, $x=-1$ is an extraneous solution.

Answer:

$x = 3$