Question

What is the solution of 4 + StartRoot 5 x + 66 EndRoot = x + 10?

Answer 1

Answer:

x = 3, answer B

Step-by-step explanation:

just took the test, putting this for anyone in the future who needs it :)

Answer 2

Answer:

Option C. x = –10 or x = 3

Step-by-step explanation:

Lets first re write clearly our expression to be solved, which reads:

$4+\sqrt{5x+66}=x+10$

Now lets first simplify it a bit as:

$\sqrt{5x+66}=x+10-4\\\sqrt{5x+66}=x+6$

Now since we have a square root on the Left Hand side we can Square both sides of the expression in order to eliminate our square root (following the rule of $(\sqrt{a})^2 =a$).

So our expression and after some required computations reads as:

$(\sqrt{5x+66})^2=(x+6)^2\\5x+66=(x+6)^2\\5x+66=x^2+12x+36\\x^2+12x-5x+36-66=0\\x^2+7x-30=0$   Eqn(1).

So now we have ended up with a typical Quadratic Equation, which in principle reads as

$ax^2+bx+c=0$ and whose roots can be found using the principle formula as:

$x_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$   Eqn(2).

So plugging in the appropriate values in Eqn(2) based on Eqn(1) we have:

$x_{1,2}=\frac{-7+/-\sqrt{7^2-4(1)(-30)} }{2(1)}\\x_{1,2}=\frac{-7+/-\sqrt{49+120} }{2}\\\\x_{1,2}=\frac{-7+/-\sqrt{169} }{2}\\\\x_{1,2}=\frac{-7+/-13}{2}$

so the two solutions for $x$ are

$x_{1}=\frac{-7-13}{2}=-10$  and $x_{2}=\frac{-7+13}{2}=3$

Thus from the options available we can say that Option C. is the correct one.