Question

What are the solutions of the quadratic equation (x – 8)2 – 13(x – 8) + 30 = 0?

x = –11 and x = –18 x = –2 and x = 5 x = 2 and x = –5 x = 11 and x = 18

Answer 1

The roots are x=18 and x=11 ..

Explanation:

Given

(x-8)^2-13(x-8)+30=0

To bring the equation in standard form

(x^2-16x+64)-13x+104+30=0

x^2-16x-13x+64+104+30=0

x^2-29x+198=0

The equation is in standard form now,

To get the solution, we have to factorize the equation.

x^2-18x-11x+198=0

x(x-18)-11(x-18)=0

(x-18)(x-11)=0

Putting the factors equal to zero

x-18=0 x-11=0

x=18 x=11

So the roots are x=18 and x=11

Answer 2

Answer: last option.

Explanation:

Remember the square of a binomial:

`(a\±b)^2=a^2\±2ab+b^2`

Given the equation
`(x-8)^2-13(x-8)+30 = 0`, you need to simplify it:

`(x^2-2(x)(8)+8^2)-13x+104+30=0\\x^2-16x+64-13x+134=0\\x^2-29x+198=0`

Use the Quadratic formula:

`x=(-b\±√(b^2-4ac))/(2a)`

In this case:

`a=1\\b=-29\\c=198`

Substituting you get:

`x=(-(-29)\±√((-29)^2-4(1)(198)))/(2(1))`

`x=11\\x=18`