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

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

Solve for x

$$x=-\frac{2}{5}=-0.4$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=20 ab=25\times 4=100$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 20.

$$a=10 b=10$$

Rewrite 25x^{2}+20x+4 as \left(25x^{2}+10x\right)+\left(10x+4\right).

$$\left(25x^{2}+10x\right)+\left(10x+4\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(5x+2\right)+2\left(5x+2\right)$$

Factor out common term 5x+2 by using distributive property.

$$\left(5x+2\right)\left(5x+2\right)$$

Rewrite as a binomial square.

$$\left(5x+2\right)^{2}$$

To find equation solution, solve 5x+2=0.

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

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 20 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 25\times 4}}{2\times 25}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 25\times 4}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-20±\sqrt{400-100\times 4}}{2\times 25}$$

Multiply -100 times 4.

$$x=\frac{-20±\sqrt{400-400}}{2\times 25}$$

Add 400 to -400.

$$x=\frac{-20±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$x=-\frac{20}{2\times 25}$$

Multiply 2 times 25.

$$x=-\frac{20}{50}$$

Reduce the fraction \frac{-20}{50} to lowest terms by extracting and canceling out 10.

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

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Subtract 4 from both sides of the equation.

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

Subtracting 4 from itself leaves 0.

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

Divide both sides by 25.

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

Dividing by 25 undoes the multiplication by 25.

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

Reduce the fraction \frac{20}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{4}{5}x=-\frac{4}{25}$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=-\frac{4}{25}+\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{-4+4}{25}$$

Add -\frac{4}{25} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=0$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{2}{5}=0 x+\frac{2}{5}=0$$

Subtract \frac{2}{5} from both sides of the equation.

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

The equation is now solved. Solutions are the same.

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