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

Question

Julia Adams · Accepted Answer

#:. x=11, x=1.# #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!

Genesis Allen · Answer

undefined To find a value of $x$ such that $f(x) = 25$ for the quadratic function $f(x) = x^2 - 12x + 36$, you need to set up the equation $f(x) = 25$ and solve for $x$.

$$
\begin{align*}
f(x) &= 25\
x^2 - 12x + 36 &= 25\
x^2 - 12x + 36 - 25 &= 0\
x^2 - 12x + 11 &= 0
\end{align*}
$$

Now, we have a quadratic equation $x^2 - 12x + 11 = 0$. To solve this equation, you can use the quadratic formula:

$$
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}
$$

For this equation, $a = 1$, $b = -12$, and $c = 11$. Substituting these values into the quadratic formula:

$$
x = \frac{{-(-12) \pm \sqrt{{(-12)^2 - 4(1)(11)}}}}{{2(1)}}
$$

$$
x = \frac{{12 \pm \sqrt{{144 - 44}}}}{{2}}
$$

$$
x = \frac{{12 \pm \sqrt{{100}}}}{{2}}
$$

$$
x = \frac{{12 \pm 10}}{{2}}
$$

This gives two possible values for $x$:

$$
x_1 = \frac{{12 + 10}}{{2}} = 11
$$

$$
x_2 = \frac{{12 - 10}}{{2}} = 1
$$

Therefore, the values of $x$ such that $f(x) = 25$ for the given quadratic function are $x = 11$ and $x = 1$.