Solve 9x^2+100x+11

$$9 x ^ { 2 } + 100 x + 11$$

Factor

$$\left(x+11\right)\left(9x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=100 ab=9\times 11=99$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 99.

$$1,99 3,33 9,11$$

Calculate the sum for each pair.

$$1+99=100 3+33=36 9+11=20$$

The solution is the pair that gives sum 100.

$$a=1 b=99$$

Rewrite 9x^{2}+100x+11 as \left(9x^{2}+x\right)+\left(99x+11\right).

$$\left(9x^{2}+x\right)+\left(99x+11\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(9x+1\right)+11\left(9x+1\right)$$

Factor out common term 9x+1 by using distributive property.

$$\left(9x+1\right)\left(x+11\right)$$

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.

$$9x^{2}+100x+11=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\times 9\times 11}}{2\times 9}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\times 9\times 11}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-100±\sqrt{10000-36\times 11}}{2\times 9}$$

Multiply -36 times 11.

$$x=\frac{-100±\sqrt{10000-396}}{2\times 9}$$

Add 10000 to -396.

$$x=\frac{-100±\sqrt{9604}}{2\times 9}$$

Take the square root of 9604.

$$x=\frac{-100±98}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-100±98}{18}$$

Now solve the equation x=\frac{-100±98}{18} when ± is plus. Add -100 to 98.

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

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

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

Now solve the equation x=\frac{-100±98}{18} when ± is minus. Subtract 98 from -100.

$$x=-\frac{198}{18}$$

Divide -198 by 18.

$$x=-11$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{9} for x_{1} and -11 for x_{2}.

$$9x^{2}+100x+11=9\left(x-\left(-\frac{1}{9}\right)\right)\left(x-\left(-11\right)\right)$$

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

$$9x^{2}+100x+11=9\left(x+\frac{1}{9}\right)\left(x+11\right)$$

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

$$9x^{2}+100x+11=9\times \frac{9x+1}{9}\left(x+11\right)$$

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

$$9x^{2}+100x+11=\left(9x+1\right)\left(x+11\right)$$

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

$$x ^ 2 +\frac{100}{9}x +\frac{11}{9} = 0$$

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

$$r + s = -\frac{100}{9} rs = \frac{11}{9}$$

Two numbers r and s sum up to -\frac{100}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{100}{9} = -\frac{50}{9}. 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>

$$r = -\frac{50}{9} - u s = -\frac{50}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{11}{9}

$$(-\frac{50}{9} - u) (-\frac{50}{9} + u) = \frac{11}{9}$$

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

$$\frac{2500}{81} - u^2 = \frac{11}{9}$$

Simplify the expression by subtracting \frac{2500}{81} on both sides

$$-u^2 = \frac{11}{9}-\frac{2500}{81} = -\frac{2401}{81}$$

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

$$u^2 = \frac{2401}{81} u = \pm\sqrt{\frac{2401}{81}} = \pm \frac{49}{9}$$

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

$$r =-\frac{50}{9} - \frac{49}{9} = -11 s = -\frac{50}{9} + \frac{49}{9} = -0.111$$

Evaluate

$$\left(x+11\right)\left(9x+1\right)$$