X 2 9x 2 0

$\exponential{(x)}{2} - 9 10 - 22 $

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a+b=-9 ab=one\left(-22\right)=-22

Factor the expression past grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-22. To find a and b, set up a system to exist solved.

ane,-22 two,-eleven

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater accented value than the positive. Listing all such integer pairs that give product -22.

1-22=-21 ii-xi=-9

Summate the sum for each pair.

a=-11 b=2

The solution is the pair that gives sum -9.

\left(x^{2}-11x\right)+\left(2x-22\correct)

Rewrite ten^{2}-9x-22 as \left(ten^{2}-11x\correct)+\left(2x-22\correct).

x\left(x-11\right)+ii\left(x-11\correct)

Factor out x in the start and two in the second grouping.

\left(x-11\right)\left(x+ii\correct)

Factor out common term x-11 by using distributive property.

x^{2}-9x-22=0

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

x=\frac{-\left(-9\correct)±\sqrt{\left(-ix\right)^{two}-iv\left(-22\right)}}{2}

All equations of the grade ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{two}-4ac}}{2a}. The quadratic formula gives ii solutions, one when ± is addition and one when information technology is subtraction.

ten=\frac{-\left(-9\correct)±\sqrt{81-4\left(-22\right)}}{2}

Square -ix.

ten=\frac{-\left(-9\right)±\sqrt{81+88}}{2}

Multiply -4 times -22.

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

Add together 81 to 88.

x=\frac{-\left(-9\correct)±13}{2}

Take the foursquare root of 169.

x=\frac{nine±xiii}{2}

The opposite of -9 is nine.

10=\frac{22}{two}

Now solve the equation x=\frac{nine±13}{two} when ± is plus. Add 9 to 13.

x=\frac{-4}{2}

Now solve the equation x=\frac{9±thirteen}{2} when ± is minus. Decrease 13 from 9.

10^{2}-9x-22=\left(10-11\right)\left(x-\left(-2\right)\correct)

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

x^{2}-9x-22=\left(ten-xi\right)\left(x+two\right)

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

x ^ 2 -9x -22 = 0

Quadratic equations such as this 1 tin can be solved by a new direct factoring method that does not crave guess work. To utilise the direct factoring method, the equation must be in the form 10^2+Bx+C=0.

r + s = 9 rs = -22

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

r = \frac{9}{ii} - u s = \frac{9}{2} + u

Two numbers r and s sum upwards to 9 exactly when the average of the two numbers is \frac{ane}{two}*9 = \frac{ix}{2}. Y'all can besides see that the midpoint of r and due south 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 southward with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.internet/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(\frac{ix}{ii} - u) (\frac{9}{ii} + u) = -22

To solve for unknown quantity u, substitute these in the product equation rs = -22

\frac{81}{4} - u^ii = -22

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

-u^2 = -22-\frac{81}{iv} = -\frac{169}{4}

Simplify the expression by subtracting \frac{81}{4} on both sides

u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}

Simplify the expression by multiplying -ane on both sides and accept the foursquare root to obtain the value of unknown variable u

r =\frac{ix}{2} - \frac{13}{2} = -ii s = \frac{9}{ii} + \frac{13}{two} = 11

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