Question

Square root of(x^2-4x-5)=2x-10

Answer 1

Answer:

x = 7

x = 5

Step-by-step explanation:

$\sqrt{x^2-4x-5}=2x-10$

We square both sides to get rid of the square root.

$(\sqrt{x^2-4x-5})^2=(2x-10)^2$

$x^2-4x-5=(2x-10)^2$

We know (a + b)^2 = a^2 + 2ab + b^2 so now the right side becomes,

$x^2-4x-5={(2x)^2-2(2x)(10)+(10)^2}$

$x^2-4x-5=4x^2-40x+100$

Now move everything to the right side,

$4x^2-x^2-40x+4x+100+5=0\\3x^2-36x+105=0$

Now we can take 3 common ,

$3(x^2-12x+35)=0\\x^2-12x+35=0$

now this is a simple quadratic equation, we can use two methods.

1) Quadratic formula

2) Factorize

We use the factorizing method,

we need two numbers when multiplied give us +35 and when added give us -12 so the numbers are -7 and -5 because -7 x -5 = +35 and

- 7 - 5 = -12

$x^2-7x-5x+35=0$

$x(x-7)-5(x-7)=0\\(x-7)(x-5)=0\\x-7=0 \ \ \ , \ \ \ x-5=0\\x=7 \ \ \ \ \ \ \ \ , \ \ \ \ x=5$