what is the solution of 4 + \\sqrt{5x + 66}=x + 10?\no x=-10\no x = 3\no x=-10 or x = 3\no no solution

Explanation:

Step1: Isolate the square - root term

Subtract 4 from both sides of the equation $4+\sqrt{5x + 66}=x + 10$. $\sqrt{5x+66}=x + 10-4$ $\sqrt{5x+66}=x + 6$

Step2: Square both sides

$(\sqrt{5x+66})^2=(x + 6)^2$ $5x+66=x^{2}+12x + 36$

Step3: Rearrange to form a quadratic equation

$x^{2}+12x+36-(5x + 66)=0$ $x^{2}+12x+36 - 5x-66=0$ $x^{2}+7x - 30=0$

Step4: Factor the quadratic equation

$x^{2}+7x - 30=(x + 10)(x - 3)=0$ So $x=-10$ or $x = 3$.

Step5: Check for extraneous solutions

When $x=-10$: Left - hand side: $\sqrt{5\times(-10)+66}+4=\sqrt{-50 + 66}+4=\sqrt{16}+4=4 + 4=8$ Right - hand side: $-10 + 10=0$ Since $8\neq0$, $x=-10$ is an extraneous solution. When $x = 3$: Left - hand side: $\sqrt{5\times3+66}+4=\sqrt{15+66}+4=\sqrt{81}+4=9 + 4=13$ Right - hand side: $3+10=13$ So $x = 3$ is the solution.

Answer:

$x = 3$