Solve for x 4x^2-20x+25=0

$4 x^{2} - 20 x + 25 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

${(2 x)}^{2} - 20 x + 25 = 0$

Rewrite $25$ as $5^{2}$ .

${(2 x)}^{2} - 20 x + 5^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$20 x = 2 \cdot (2 x) \cdot 5$

Rewrite the polynomial.

${(2 x)}^{2} - 2 \cdot (2 x) \cdot 5 + 5^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 5$ .

${(2 x - 5)}^{2} = 0$

Step 2

Set the $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 3

Solve for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2}$

Decimal Form:

$x = 2.5$

Mixed Number Form:

$x = 2 \frac{1}{2}$