Answer
$x=\dfrac{5}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Express the given equation, $
4x^2-20x+25=0
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
(2x-5)(2x-5)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
2x-5=0
\text{ OR }
2x-5=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
2x-5=0
\\\\
2x=5
\\\\
x=\dfrac{5}{2}
\\\\\text{ OR }\\\\
2x-5=0
\\\\
2x=5
\\\\
x=\dfrac{5}{2}
.\end{array}
Hence, the solutions are $
x=\dfrac{5}{2}
.$
$\bf{\text{Supplementary Solution/s:}}$
To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping.
In the expression, $
4x^2-20x+25
,$ the value of $ac$ is $
4(25)=100
$ and the value of $b$ is $
-20
.$
The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{
-10,-10
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
4x^2-10x-10x+25
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(4x^2-10x)-(10x-25)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2x(2x-5)-5(2x-5)
.\end{array}
Factoring the $GCF=
(2x-5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2x-5)(2x-5)
.\end{array}
