solve the following radical equation. $sqrt{3x + 15}+3=x + 2$ answer how to enter your answer (opens in new…

Explanation:

Step1: Isolate the radical term

Subtract 3 from both sides of the equation $\sqrt{3x + 15}+3=x + 2$. We get $\sqrt{3x+15}=x - 1$.

Step2: Square both sides

Squaring both sides of $\sqrt{3x+15}=x - 1$, we have $(\sqrt{3x+15})^2=(x - 1)^2$. Using the formula $(a - b)^2=a^2-2ab + b^2$, we obtain $3x+15=x^2-2x + 1$.

Step3: Rearrange to a quadratic - equation

Rearrange the equation $3x+15=x^2-2x + 1$ to the standard quadratic form $ax^2+bx + c = 0$. $x^2-2x-3x+1 - 15=0$, which simplifies to $x^2-5x-14 = 0$.

Step4: Factor the quadratic equation

Factor $x^2-5x-14 = 0$. We need two numbers that multiply to - 14 and add up to - 5. The numbers are - 7 and 2. So, $(x - 7)(x+2)=0$.

Step5: Solve for x

Set each factor equal to zero: If $x - 7=0$, then $x = 7$; if $x+2=0$, then $x=-2$.

Step6: Check for extraneous solutions

Substitute $x = 7$ into the original equation: Left - hand side: $\sqrt{3\times7+15}+3=\sqrt{21 + 15}+3=\sqrt{36}+3=6 + 3=9$. Right - hand side: $7 + 2=9$. So, $x = 7$ is a valid solution. Substitute $x=-2$ into the original equation: Left - hand side: $\sqrt{3\times(-2)+15}+3=\sqrt{-6 + 15}+3=\sqrt{9}+3=3 + 3=6$. Right - hand side: $-2+2=0$. Since $6\neq0$, $x=-2$ is an extraneous solution.

Answer:

$x = 7$