Solve 4x^2-100x+225=0

$$4 { x }^{ 2 } -100x+225=0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = \frac{45}{2} = 22\frac{1}{2} = 22.5$$

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 4x^{2}+ax+bx+225. To find a and b, set up a system to be solved.

$$a+b=-100 ab=4\times 225=900$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 900.

$$-1,-900 -2,-450 -3,-300 -4,-225 -5,-180 -6,-150 -9,-100 -10,-90 -12,-75 -15,-60 -18,-50 -20,-45 -25,-36 -30,-30$$

Calculate the sum for each pair.

$$-1-900=-901 -2-450=-452 -3-300=-303 -4-225=-229 -5-180=-185 -6-150=-156 -9-100=-109 -10-90=-100 -12-75=-87 -15-60=-75 -18-50=-68 -20-45=-65 -25-36=-61 -30-30=-60$$

The solution is the pair that gives sum -100.

$$a=-90 b=-10$$

Rewrite 4x^{2}-100x+225 as \left(4x^{2}-90x\right)+\left(-10x+225\right).

$$\left(4x^{2}-90x\right)+\left(-10x+225\right)$$

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

$$2x\left(2x-45\right)-5\left(2x-45\right)$$

Factor out common term 2x-45 by using distributive property.

$$\left(2x-45\right)\left(2x-5\right)$$

To find equation solutions, solve 2x-45=0 and 2x-5=0.

$$x=\frac{45}{2} x=\frac{5}{2}$$

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.

$$4x^{2}-100x+225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -100 for b, and 225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 4\times 225}}{2\times 4}$$

Square -100.

$$x=\frac{-\left(-100\right)±\sqrt{10000-4\times 4\times 225}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-100\right)±\sqrt{10000-16\times 225}}{2\times 4}$$

Multiply -16 times 225.

$$x=\frac{-\left(-100\right)±\sqrt{10000-3600}}{2\times 4}$$

Add 10000 to -3600.

$$x=\frac{-\left(-100\right)±\sqrt{6400}}{2\times 4}$$

Take the square root of 6400.

$$x=\frac{-\left(-100\right)±80}{2\times 4}$$

The opposite of -100 is 100.

$$x=\frac{100±80}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{100±80}{8}$$

Now solve the equation x=\frac{100±80}{8} when ± is plus. Add 100 to 80.

$$x=\frac{180}{8}$$

Reduce the fraction \frac{180}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{45}{2}$$

Now solve the equation x=\frac{100±80}{8} when ± is minus. Subtract 80 from 100.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{45}{2} x=\frac{5}{2}$$

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.

$$4x^{2}-100x+225=0$$

Subtract 225 from both sides of the equation.

$$4x^{2}-100x+225-225=-225$$

Subtracting 225 from itself leaves 0.

$$4x^{2}-100x=-225$$

Divide both sides by 4.

$$\frac{4x^{2}-100x}{4}=-\frac{225}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{100}{4}\right)x=-\frac{225}{4}$$

Divide -100 by 4.

$$x^{2}-25x=-\frac{225}{4}$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-\frac{225}{4}+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=\frac{-225+625}{4}$$

Add -\frac{225}{4} to \frac{625}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-25x+\frac{625}{4}=100$$

Factor x^{2}-25x+\frac{625}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{2}\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{100}$$

Simplify.

$$x-\frac{25}{2}=10 x-\frac{25}{2}=-10$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{45}{2} x=\frac{5}{2}$$