Solve 16x^2=40x-9

$$16 x ^ { 2 } = 40 x - 9$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Steps Using Factoring By Grouping

Subtract 40x from both sides.

$$16x^{2}-40x=-9$$

Add 9 to both sides.

$$16x^{2}-40x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-40 ab=16\times 9=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -40.

$$a=-36 b=-4$$

Rewrite 16x^{2}-40x+9 as \left(16x^{2}-36x\right)+\left(-4x+9\right).

$$\left(16x^{2}-36x\right)+\left(-4x+9\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(4x-9\right)-\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(4x-1\right)$$

To find equation solutions, solve 4x-9=0 and 4x-1=0.

$$x=\frac{9}{4} x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Subtract 40x from both sides.

$$16x^{2}-40x=-9$$

Add 9 to both sides.

$$16x^{2}-40x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -40 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 16\times 9}}{2\times 16}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 16\times 9}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-40\right)±\sqrt{1600-64\times 9}}{2\times 16}$$

Multiply -64 times 9.

$$x=\frac{-\left(-40\right)±\sqrt{1600-576}}{2\times 16}$$

Add 1600 to -576.

$$x=\frac{-\left(-40\right)±\sqrt{1024}}{2\times 16}$$

Take the square root of 1024.

$$x=\frac{-\left(-40\right)±32}{2\times 16}$$

The opposite of -40 is 40.

$$x=\frac{40±32}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{40±32}{32}$$

Now solve the equation x=\frac{40±32}{32} when ± is plus. Add 40 to 32.

$$x=\frac{72}{32}$$

Reduce the fraction \frac{72}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{40±32}{32} when ± is minus. Subtract 32 from 40.

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

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{4}$$

The equation is now solved.

$$x=\frac{9}{4} x=\frac{1}{4}$$

Steps for Completing the Square

Subtract 40x from both sides.

$$16x^{2}-40x=-9$$

Divide both sides by 16.

$$\frac{16x^{2}-40x}{16}=-\frac{9}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{40}{16}\right)x=-\frac{9}{16}$$

Reduce the fraction \frac{-40}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{5}{2}x=-\frac{9}{16}$$

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

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{9}{16}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{-9+25}{16}$$

Add -\frac{9}{16} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=1$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-\frac{5}{4}=1 x-\frac{5}{4}=-1$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{9}{4} x=\frac{1}{4}$$