Solve 2j^2-9j+9

$$2 j ^ { 2 } - 9 j + 9$$

Factor

$$\left(j-3\right)\left(2j-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2j^{2}+aj+bj+9. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\times 9=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite 2j^{2}-9j+9 as \left(2j^{2}-6j\right)+\left(-3j+9\right).

$$\left(2j^{2}-6j\right)+\left(-3j+9\right)$$

Factor out 2j in the first and -3 in the second group.

$$2j\left(j-3\right)-3\left(j-3\right)$$

Factor out common term j-3 by using distributive property.

$$\left(j-3\right)\left(2j-3\right)$$

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.

$$2j^{2}-9j+9=0$$

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.

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

Square -9.

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

Multiply -4 times 2.

$$j=\frac{-\left(-9\right)±\sqrt{81-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$j=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 2}$$

Add 81 to -72.

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

Take the square root of 9.

$$j=\frac{-\left(-9\right)±3}{2\times 2}$$

The opposite of -9 is 9.

$$j=\frac{9±3}{2\times 2}$$

Multiply 2 times 2.

$$j=\frac{9±3}{4}$$

Now solve the equation j=\frac{9±3}{4} when ± is plus. Add 9 to 3.

$$j=\frac{12}{4}$$

Divide 12 by 4.

$$j=3$$

Now solve the equation j=\frac{9±3}{4} when ± is minus. Subtract 3 from 9.

$$j=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

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

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

Subtract \frac{3}{2} from j by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 2, the greatest common factor in 2 and 2.

$$2j^{2}-9j+9=\left(j-3\right)\left(2j-3\right)$$

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 2

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

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

$$r + s = \frac{9}{2} rs = \frac{9}{2}$$

Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{4}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}

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

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

$$\frac{81}{16} - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = \frac{9}{2}-\frac{81}{16} = -\frac{9}{16}$$

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

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

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

$$r =\frac{9}{4} - \frac{3}{4} = 1.500 s = \frac{9}{4} + \frac{3}{4} = 3$$

Evaluate

$$\left(j-3\right)\left(2j-3\right)$$