Solve 9(1-4x)^2+5=30

$$9 ( 1 - 4 x ) ^ { 2 } + 5 = 30$$

Solve for x

$$x=-\frac{1}{6}\approx -0.166666667$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-4x\right)^{2}.

$$9\left(1-8x+16x^{2}\right)+5=30$$

Use the distributive property to multiply 9 by 1-8x+16x^{2}.

$$9-72x+144x^{2}+5=30$$

Add 9 and 5 to get 14.

$$14-72x+144x^{2}=30$$

Subtract 30 from both sides.

$$14-72x+144x^{2}-30=0$$

Subtract 30 from 14 to get -16.

$$-16-72x+144x^{2}=0$$

Divide both sides by 8.

$$-2-9x+18x^{2}=0$$

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

$$18x^{2}-9x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 18x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=-9 ab=18\left(-2\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$a=-12 b=3$$

Rewrite 18x^{2}-9x-2 as \left(18x^{2}-12x\right)+\left(3x-2\right).

$$\left(18x^{2}-12x\right)+\left(3x-2\right)$$

Factor out 6x in 18x^{2}-12x.

$$6x\left(3x-2\right)+3x-2$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(6x+1\right)$$

To find equation solutions, solve 3x-2=0 and 6x+1=0.

$$x=\frac{2}{3} x=-\frac{1}{6}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-4x\right)^{2}.

$$9\left(1-8x+16x^{2}\right)+5=30$$

Use the distributive property to multiply 9 by 1-8x+16x^{2}.

$$9-72x+144x^{2}+5=30$$

Add 9 and 5 to get 14.

$$14-72x+144x^{2}=30$$

Subtract 30 from both sides.

$$14-72x+144x^{2}-30=0$$

Subtract 30 from 14 to get -16.

$$-16-72x+144x^{2}=0$$

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.

$$144x^{2}-72x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 144 for a, -72 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-72\right)±\sqrt{\left(-72\right)^{2}-4\times 144\left(-16\right)}}{2\times 144}$$

Square -72.

$$x=\frac{-\left(-72\right)±\sqrt{5184-4\times 144\left(-16\right)}}{2\times 144}$$

Multiply -4 times 144.

$$x=\frac{-\left(-72\right)±\sqrt{5184-576\left(-16\right)}}{2\times 144}$$

Multiply -576 times -16.

$$x=\frac{-\left(-72\right)±\sqrt{5184+9216}}{2\times 144}$$

Add 5184 to 9216.

$$x=\frac{-\left(-72\right)±\sqrt{14400}}{2\times 144}$$

Take the square root of 14400.

$$x=\frac{-\left(-72\right)±120}{2\times 144}$$

The opposite of -72 is 72.

$$x=\frac{72±120}{2\times 144}$$

Multiply 2 times 144.

$$x=\frac{72±120}{288}$$

Now solve the equation x=\frac{72±120}{288} when ± is plus. Add 72 to 120.

$$x=\frac{192}{288}$$

Reduce the fraction \frac{192}{288} to lowest terms by extracting and canceling out 96.

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

Now solve the equation x=\frac{72±120}{288} when ± is minus. Subtract 120 from 72.

$$x=-\frac{48}{288}$$

Reduce the fraction \frac{-48}{288} to lowest terms by extracting and canceling out 48.

$$x=-\frac{1}{6}$$

The equation is now solved.

$$x=\frac{2}{3} x=-\frac{1}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-4x\right)^{2}.

$$9\left(1-8x+16x^{2}\right)+5=30$$

Use the distributive property to multiply 9 by 1-8x+16x^{2}.

$$9-72x+144x^{2}+5=30$$

Add 9 and 5 to get 14.

$$14-72x+144x^{2}=30$$

Subtract 14 from both sides.

$$-72x+144x^{2}=30-14$$

Subtract 14 from 30 to get 16.

$$-72x+144x^{2}=16$$

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.

$$144x^{2}-72x=16$$

Divide both sides by 144.

$$\frac{144x^{2}-72x}{144}=\frac{16}{144}$$

Dividing by 144 undoes the multiplication by 144.

$$x^{2}+\left(-\frac{72}{144}\right)x=\frac{16}{144}$$

Reduce the fraction \frac{-72}{144} to lowest terms by extracting and canceling out 72.

$$x^{2}-\frac{1}{2}x=\frac{16}{144}$$

Reduce the fraction \frac{16}{144} to lowest terms by extracting and canceling out 16.

$$x^{2}-\frac{1}{2}x=\frac{1}{9}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{9}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{9}+\frac{1}{16}$$

Add \frac{1}{9} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{144}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$x-\frac{1}{4}=\frac{5}{12} x-\frac{1}{4}=-\frac{5}{12}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{2}{3} x=-\frac{1}{6}$$