Given the quadratic function f(x) = x^2 - 12x + 36 how do you find a value of x such that f(x) = 25?

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Given the quadratic function f(x) = x^2 - 12x + 36 how do you find a value of x such that f(x) = 25?

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Answer 1 · 2016-06-20T18:25:08

$:. x=11, x=1.$

Explanation:

$f(x)=25$
$:. x^2-12x+36=25.$
$:. x^2-12x+36-25=0.$
$:. x^2-12x+11=0.$
$:. x^2-11x-1x+11=0.$ ......... $[11xx1=11,11+1=12]$
$:. x(x-11)-1(x-11)=0.$
$:. (x-11)(x-1)=0.$
$:. x=11, x=1.$

Alternatively, we see that $f(x)=(x-6)^2.$
Hence, $f(x)=25$
$rArr (x-6)^2=25=5^2.$
$rArr (x-6)=+-5$
$rArr x=6+-5$
$rArr x=11, or, x=1,$ as before!

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Given the quadratic function f(x) = x^2 - 12x + 36 how do you find a value of x such that f(x) = 25?

1 Answer

Explanation:

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