Solve 16-9x^2=(x-1)(3x-4)

$$16 - 9 x ^ { 2 } = ( x - 1 ) ( 3 x - 4 )$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-1 by 3x-4 and combine like terms.

$$16-9x^{2}=3x^{2}-7x+4$$

Subtract 3x^{2} from both sides.

$$16-9x^{2}-3x^{2}=-7x+4$$

Combine -9x^{2} and -3x^{2} to get -12x^{2}.

$$16-12x^{2}=-7x+4$$

Add 7x to both sides.

$$16-12x^{2}+7x=4$$

Subtract 4 from both sides.

$$16-12x^{2}+7x-4=0$$

Subtract 4 from 16 to get 12.

$$12-12x^{2}+7x=0$$

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

$$-12x^{2}+7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -12x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=7 ab=-12\times 12=-144$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the 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.

$$-1+144=143 -2+72=70 -3+48=45 -4+36=32 -6+24=18 -8+18=10 -9+16=7 -12+12=0$$

The solution is the pair that gives sum 7.

$$a=16 b=-9$$

Rewrite -12x^{2}+7x+12 as \left(-12x^{2}+16x\right)+\left(-9x+12\right).

$$\left(-12x^{2}+16x\right)+\left(-9x+12\right)$$

Factor out -4x in the first and -3 in the second group.

$$-4x\left(3x-4\right)-3\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(-4x-3\right)$$

To find equation solutions, solve 3x-4=0 and -4x-3=0.

$$x=\frac{4}{3} x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-1 by 3x-4 and combine like terms.

$$16-9x^{2}=3x^{2}-7x+4$$

Subtract 3x^{2} from both sides.

$$16-9x^{2}-3x^{2}=-7x+4$$

Combine -9x^{2} and -3x^{2} to get -12x^{2}.

$$16-12x^{2}=-7x+4$$

Add 7x to both sides.

$$16-12x^{2}+7x=4$$

Subtract 4 from both sides.

$$16-12x^{2}+7x-4=0$$

Subtract 4 from 16 to get 12.

$$12-12x^{2}+7x=0$$

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.

$$-12x^{2}+7x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-12\right)\times 12}}{2\left(-12\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-12\right)\times 12}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-7±\sqrt{49+48\times 12}}{2\left(-12\right)}$$

Multiply 48 times 12.

$$x=\frac{-7±\sqrt{49+576}}{2\left(-12\right)}$$

Add 49 to 576.

$$x=\frac{-7±\sqrt{625}}{2\left(-12\right)}$$

Take the square root of 625.

$$x=\frac{-7±25}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-7±25}{-24}$$

Now solve the equation x=\frac{-7±25}{-24} when ± is plus. Add -7 to 25.

$$x=\frac{18}{-24}$$

Reduce the fraction \frac{18}{-24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{3}{4}$$

Now solve the equation x=\frac{-7±25}{-24} when ± is minus. Subtract 25 from -7.

$$x=-\frac{32}{-24}$$

Reduce the fraction \frac{-32}{-24} to lowest terms by extracting and canceling out 8.

$$x=\frac{4}{3}$$

The equation is now solved.

$$x=-\frac{3}{4} x=\frac{4}{3}$$

Steps for Completing the Square

Use the distributive property to multiply x-1 by 3x-4 and combine like terms.

$$16-9x^{2}=3x^{2}-7x+4$$

Subtract 3x^{2} from both sides.

$$16-9x^{2}-3x^{2}=-7x+4$$

Combine -9x^{2} and -3x^{2} to get -12x^{2}.

$$16-12x^{2}=-7x+4$$

Add 7x to both sides.

$$16-12x^{2}+7x=4$$

Subtract 16 from both sides.

$$-12x^{2}+7x=4-16$$

Subtract 16 from 4 to get -12.

$$-12x^{2}+7x=-12$$

Divide both sides by -12.

$$\frac{-12x^{2}+7x}{-12}=-\frac{12}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{7}{-12}x=-\frac{12}{-12}$$

Divide 7 by -12.

$$x^{2}-\frac{7}{12}x=-\frac{12}{-12}$$

Divide -12 by -12.

$$x^{2}-\frac{7}{12}x=1$$

Divide -\frac{7}{12}, the coefficient of the x term, by 2 to get -\frac{7}{24}. Then add the square of -\frac{7}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{12}x+\left(-\frac{7}{24}\right)^{2}=1+\left(-\frac{7}{24}\right)^{2}$$

Square -\frac{7}{24} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{12}x+\frac{49}{576}=1+\frac{49}{576}$$

Add 1 to \frac{49}{576}.

$$x^{2}-\frac{7}{12}x+\frac{49}{576}=\frac{625}{576}$$

Factor x^{2}-\frac{7}{12}x+\frac{49}{576}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{24}\right)^{2}=\frac{625}{576}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{24}\right)^{2}}=\sqrt{\frac{625}{576}}$$

Simplify.

$$x-\frac{7}{24}=\frac{25}{24} x-\frac{7}{24}=-\frac{25}{24}$$

Add \frac{7}{24} to both sides of the equation.

$$x=\frac{4}{3} x=-\frac{3}{4}$$