Solve (3x-2)(3x+2)=35x

$$( 3 x - 2 ) ( 3 x + 2 ) = 35 x$$

Solve for x

$$x=-\frac{1}{9}\approx -0.111111111$$

$$x=4$$

Steps Using Factoring By Grouping

Consider \left(3x-2\right)\left(3x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$\left(3x\right)^{2}-4=35x$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4=35x$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4=35x$$

Subtract 35x from both sides.

$$9x^{2}-4-35x=0$$

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

$$9x^{2}-35x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-35 ab=9\left(-4\right)=-36$$

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 -35.

$$a=-36 b=1$$

Rewrite 9x^{2}-35x-4 as \left(9x^{2}-36x\right)+\left(x-4\right).

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

Factor out 9x in 9x^{2}-36x.

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

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

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

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

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

Steps Using the Quadratic Formula

Consider \left(3x-2\right)\left(3x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$\left(3x\right)^{2}-4=35x$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4=35x$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4=35x$$

Subtract 35x from both sides.

$$9x^{2}-4-35x=0$$

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.

$$9x^{2}-35x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -35 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-35\right)±\sqrt{1225-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$x=\frac{-\left(-35\right)±\sqrt{1225+144}}{2\times 9}$$

Add 1225 to 144.

$$x=\frac{-\left(-35\right)±\sqrt{1369}}{2\times 9}$$

Take the square root of 1369.

$$x=\frac{-\left(-35\right)±37}{2\times 9}$$

The opposite of -35 is 35.

$$x=\frac{35±37}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{35±37}{18}$$

Now solve the equation x=\frac{35±37}{18} when ± is plus. Add 35 to 37.

$$x=\frac{72}{18}$$

Divide 72 by 18.

$$x=4$$

Now solve the equation x=\frac{35±37}{18} when ± is minus. Subtract 37 from 35.

$$x=-\frac{2}{18}$$

Reduce the fraction \frac{-2}{18} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Consider \left(3x-2\right)\left(3x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$\left(3x\right)^{2}-4=35x$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4=35x$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4=35x$$

Subtract 35x from both sides.

$$9x^{2}-4-35x=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$9x^{2}-35x=4$$

Divide both sides by 9.

$$\frac{9x^{2}-35x}{9}=\frac{4}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{35}{9}x=\frac{4}{9}$$

Divide -\frac{35}{9}, the coefficient of the x term, by 2 to get -\frac{35}{18}. Then add the square of -\frac{35}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{35}{9}x+\left(-\frac{35}{18}\right)^{2}=\frac{4}{9}+\left(-\frac{35}{18}\right)^{2}$$

Square -\frac{35}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{35}{9}x+\frac{1225}{324}=\frac{4}{9}+\frac{1225}{324}$$

Add \frac{4}{9} to \frac{1225}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{35}{9}x+\frac{1225}{324}=\frac{1369}{324}$$

Factor x^{2}-\frac{35}{9}x+\frac{1225}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{35}{18}\right)^{2}=\frac{1369}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{35}{18}\right)^{2}}=\sqrt{\frac{1369}{324}}$$

Simplify.

$$x-\frac{35}{18}=\frac{37}{18} x-\frac{35}{18}=-\frac{37}{18}$$

Add \frac{35}{18} to both sides of the equation.

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