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a+b=-9 ab=-52
To solve the equation, factor x^{2}-9x-52 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
1,-52 2,-26 4,-13
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -52.
1-52=-51 2-26=-24 4-13=-9
Calculate the sum for each pair.
a=-13 b=4
The solution is the pair that gives sum -9.
\left(x-13\right)\left(x+4\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=13 x=-4
To find equation solutions, solve x-13=0 and x+4=0.
a+b=-9 ab=1\left(-52\right)=-52
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-52. To find a and b, set up a system to be solved.
1,-52 2,-26 4,-13
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -52.
1-52=-51 2-26=-24 4-13=-9
Calculate the sum for each pair.
a=-13 b=4
The solution is the pair that gives sum -9.
\left(x^{2}-13x\right)+\left(4x-52\right)
Rewrite x^{2}-9x-52 as \left(x^{2}-13x\right)+\left(4x-52\right).
x\left(x-13\right)+4\left(x-13\right)
Factor out x in the first and 4 in the second group.
\left(x-13\right)\left(x+4\right)
Factor out common term x-13 by using distributive property.
x=13 x=-4
To find equation solutions, solve x-13=0 and x+4=0.
x^{2}-9x-52=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{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-52\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and -52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-9\right)±\sqrt{81-4\left(-52\right)}}{2}
Square -9.
x=\frac{-\left(-9\right)±\sqrt{81+208}}{2}
Multiply -4 times -52.
x=\frac{-\left(-9\right)±\sqrt{289}}{2}
Add 81 to 208.
x=\frac{-\left(-9\right)±17}{2}
Take the square root of 289.
x=\frac{9±17}{2}
The opposite of -9 is 9.
x=\frac{26}{2}
Now solve the equation x=\frac{9±17}{2} when ± is plus. Add 9 to 17.
x=13
Divide 26 by 2.
x=-\frac{8}{2}
Now solve the equation x=\frac{9±17}{2} when ± is minus. Subtract 17 from 9.
x=-4
Divide -8 by 2.
x=13 x=-4
The equation is now solved.
x^{2}-9x-52=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
x^{2}-9x-52-\left(-52\right)=-\left(-52\right)
Add 52 to both sides of the equation.
x^{2}-9x=-\left(-52\right)
Subtracting -52 from itself leaves 0.
x^{2}-9x=52
Subtract -52 from 0.
x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=52+\left(-\frac{9}{2}\right)^{2}
Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-9x+\frac{81}{4}=52+\frac{81}{4}
Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-9x+\frac{81}{4}=\frac{289}{4}
Add 52 to \frac{81}{4}.
\left(x-\frac{9}{2}\right)^{2}=\frac{289}{4}
Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{289}{4}}
Take the square root of both sides of the equation.
x-\frac{9}{2}=\frac{17}{2} x-\frac{9}{2}=-\frac{17}{2}
Simplify.
x=13 x=-4
Add \frac{9}{2} to both sides of the equation.
x ^ 2 -9x -52 = 0
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.
r + s = 9 rs = -52
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 = \frac{9}{2} - u s = \frac{9}{2} + u
Two numbers r and s sum up to 9 exactly when the average of the two numbers is \frac{1}{2}*9 = \frac{9}{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>
(\frac{9}{2} - u) (\frac{9}{2} + u) = -52
To solve for unknown quantity u, substitute these in the product equation rs = -52
\frac{81}{4} - u^2 = -52
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -52-\frac{81}{4} = -\frac{289}{4}
Simplify the expression by subtracting \frac{81}{4} on both sides
u^2 = \frac{289}{4} u = \pm\sqrt{\frac{289}{4}} = \pm \frac{17}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{9}{2} - \frac{17}{2} = -4 s = \frac{9}{2} + \frac{17}{2} = 13
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.