Solve left(3xright)^2-13x+4=0

$${ \left(3x \right) }^{ 2 } -13x+4 = 0$$

Solve for x

$$x=1$$

$$x=\frac{4}{9}\approx 0.444444444$$

Steps Using Factoring By Grouping

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

$$3^{2}x^{2}-13x+4=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-13x+4=0$$

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.

$$a+b=-13 ab=9\times 4=36$$

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 36.

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

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -13.

$$a=-9 b=-4$$

Rewrite 9x^{2}-13x+4 as \left(9x^{2}-9x\right)+\left(-4x+4\right).

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

Factor out 9x in the first and -4 in the second group.

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

Factor out common term x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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

$$3^{2}x^{2}-13x+4=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-13x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -13 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 9\times 4}}{2\times 9}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 9\times 4}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-13\right)±\sqrt{169-36\times 4}}{2\times 9}$$

Multiply -36 times 4.

$$x=\frac{-\left(-13\right)±\sqrt{169-144}}{2\times 9}$$

Add 169 to -144.

$$x=\frac{-\left(-13\right)±\sqrt{25}}{2\times 9}$$

Take the square root of 25.

$$x=\frac{-\left(-13\right)±5}{2\times 9}$$

The opposite of -13 is 13.

$$x=\frac{13±5}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{13±5}{18}$$

Now solve the equation x=\frac{13±5}{18} when ± is plus. Add 13 to 5.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{13±5}{18} when ± is minus. Subtract 5 from 13.

$$x=\frac{8}{18}$$

Reduce the fraction \frac{8}{18} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$3^{2}x^{2}-13x+4=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-13x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-13x=-4$$

Divide both sides by 9.

$$\frac{9x^{2}-13x}{9}=-\frac{4}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{13}{9}x=-\frac{4}{9}$$

Divide -\frac{13}{9}, the coefficient of the x term, by 2 to get -\frac{13}{18}. Then add the square of -\frac{13}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{9}x+\left(-\frac{13}{18}\right)^{2}=-\frac{4}{9}+\left(-\frac{13}{18}\right)^{2}$$

Square -\frac{13}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{9}x+\frac{169}{324}=-\frac{4}{9}+\frac{169}{324}$$

Add -\frac{4}{9} to \frac{169}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{13}{9}x+\frac{169}{324}=\frac{25}{324}$$

Factor x^{2}-\frac{13}{9}x+\frac{169}{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}.

$$\left(x-\frac{13}{18}\right)^{2}=\frac{25}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{18}\right)^{2}}=\sqrt{\frac{25}{324}}$$

Simplify.

$$x-\frac{13}{18}=\frac{5}{18} x-\frac{13}{18}=-\frac{5}{18}$$

Add \frac{13}{18} to both sides of the equation.

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