Solve 9x^2+9x-4

$$9 x ^ { 2 } + 9 x - 4$$

Factor

$$\left(3x-1\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

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

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 -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.

$$a=-3 b=12$$

Rewrite 9x^{2}+9x-4 as \left(9x^{2}-3x\right)+\left(12x-4\right).

$$\left(9x^{2}-3x\right)+\left(12x-4\right)$$

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

$$3x\left(3x-1\right)+4\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(3x+4\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

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.

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

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-9±\sqrt{81-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$x=\frac{-9±\sqrt{81+144}}{2\times 9}$$

Add 81 to 144.

$$x=\frac{-9±\sqrt{225}}{2\times 9}$$

Take the square root of 225.

$$x=\frac{-9±15}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-9±15}{18}$$

Now solve the equation x=\frac{-9±15}{18} when ± is plus. Add -9 to 15.

$$x=\frac{6}{18}$$

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-9±15}{18} when ± is minus. Subtract 15 from -9.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{3} for x_{1} and -\frac{4}{3} for x_{2}.

$$9x^{2}+9x-4=9\left(x-\frac{1}{3}\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$9x^{2}+9x-4=9\left(x-\frac{1}{3}\right)\left(x+\frac{4}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+9x-4=9\times \frac{3x-1}{3}\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+9x-4=9\times \frac{3x-1}{3}\times \frac{3x+4}{3}$$

Multiply \frac{3x-1}{3} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+9x-4=9\times \frac{\left(3x-1\right)\left(3x+4\right)}{3\times 3}$$

Multiply 3 times 3.

$$9x^{2}+9x-4=9\times \frac{\left(3x-1\right)\left(3x+4\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}+9x-4=\left(3x-1\right)\left(3x+4\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 9

$$x ^ 2 +1x -\frac{4}{9} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -1 rs = -\frac{4}{9}$$

Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{9}

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -\frac{4}{9}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1}{4} - u^2 = -\frac{4}{9}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{4}{9}-\frac{1}{4} = -\frac{25}{36}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{1}{2} - \frac{5}{6} = -1.333 s = -\frac{1}{2} + \frac{5}{6} = 0.333$$

Evaluate

$$\left(3x-1\right)\left(3x+4\right)$$