Question

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

Answer 1

Answer: last option.

Step-by-step 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:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

In this case:

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

Substituting you get:

$x=\frac{-(-29)\±\sqrt{(-29)^2-4(1)(198)}}{2(1)}$

$x=11\\x=18$

Answer 2

Answer:

The roots are x=18 and x=11  ..

Step-by-step 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