Solve 4^2-4x(5-x)=0

$${ 4 }^{ 2 } -4x(5-x)=0$$

Solve for x

$$x=1$$

$$x=4$$

Steps Using Factoring By Grouping

Calculate 4 to the power of 2 and get 16.

$$16-4x\left(5-x\right)=0$$

Use the distributive property to multiply -4x by 5-x.

$$16-20x+4x^{2}=0$$

Divide both sides by 4.

$$4-5x+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-5x+4=0$$

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

$$a+b=-5 ab=1\times 4=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-1\right)$$

To find equation solutions, solve x-4=0 and x-1=0.

$$x=4 x=1$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16-4x\left(5-x\right)=0$$

Use the distributive property to multiply -4x by 5-x.

$$16-20x+4x^{2}=0$$

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.

$$4x^{2}-20x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 16}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 16}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 16}}{2\times 4}$$

Multiply -16 times 16.

$$x=\frac{-\left(-20\right)±\sqrt{400-256}}{2\times 4}$$

Add 400 to -256.

$$x=\frac{-\left(-20\right)±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{-\left(-20\right)±12}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±12}{8}$$

Now solve the equation x=\frac{20±12}{8} when ± is plus. Add 20 to 12.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

Now solve the equation x=\frac{20±12}{8} when ± is minus. Subtract 12 from 20.

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

Divide 8 by 8.

$$x=1$$

The equation is now solved.

$$x=4 x=1$$

Steps for Completing the Square

Calculate 4 to the power of 2 and get 16.

$$16-4x\left(5-x\right)=0$$

Use the distributive property to multiply -4x by 5-x.

$$16-20x+4x^{2}=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-20x+4x^{2}=-16$$

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.

$$4x^{2}-20x=-16$$

Divide both sides by 4.

$$\frac{4x^{2}-20x}{4}=-\frac{16}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{16}{4}$$

Divide -20 by 4.

$$x^{2}-5x=-\frac{16}{4}$$

Divide -16 by 4.

$$x^{2}-5x=-4$$

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

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

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

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

Add -4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=4 x=1$$