Answer: last option.
Step-by-step explanation:
Remember the square of a binomial:
[tex](a\±b)^2=a^2\±2ab+b^2[/tex]
Given the equation [tex](x-8)^2-13(x-8)+30 = 0[/tex], you need to simplify it:
[tex](x^2-2(x)(8)+8^2)-13x+104+30=0\\x^2-16x+64-13x+134=0\\x^2-29x+198=0[/tex]
Use the Quadratic formula:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
In this case:
[tex]a=1\\b=-29\\c=198[/tex]
Substituting you get:
[tex]x=\frac{-(-29)\±\sqrt{(-29)^2-4(1)(198)}}{2(1)}[/tex]
[tex]x=11\\x=18[/tex]
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
