$\exponential{x}{2} + x - 56 = 0 $

Solve for x

x=-8

x=7

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a+b=1 ab=-56

To solve the equation, factor x^{2}+x-56 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.

-1,56 -2,28 -4,14 -7,8

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 -56.

-1+56=55 -2+28=26 -4+14=10 -7+8=1

Calculate the sum for each pair.

a=-7 b=8

The solution is the pair that gives sum 1.

\left(x-7\right)\left(x+8\right)

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

x=7 x=-8

To find equation solutions, solve x-7=0 and x+8=0.

a+b=1 ab=1\left(-56\right)=-56

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-56. To find a and b, set up a system to be solved.

-1,56 -2,28 -4,14 -7,8

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 -56.

-1+56=55 -2+28=26 -4+14=10 -7+8=1

Calculate the sum for each pair.

a=-7 b=8

The solution is the pair that gives sum 1.

\left(x^{2}-7x\right)+\left(8x-56\right)

Rewrite x^{2}+x-56 as \left(x^{2}-7x\right)+\left(8x-56\right).

x\left(x-7\right)+8\left(x-7\right)

Factor out x in the first and 8 in the second group.

\left(x-7\right)\left(x+8\right)

Factor out common term x-7 by using distributive property.

x=7 x=-8

To find equation solutions, solve x-7=0 and x+8=0.

x^{2}+x-56=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{-1±\sqrt{1^{2}-4\left(-56\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-1±\sqrt{1-4\left(-56\right)}}{2}

Square 1.

x=\frac{-1±\sqrt{1+224}}{2}

Multiply -4 times -56.

x=\frac{-1±\sqrt{225}}{2}

Add 1 to 224.

x=\frac{-1±15}{2}

Take the square root of 225.

x=\frac{14}{2}

Now solve the equation x=\frac{-1±15}{2} when ± is plus. Add -1 to 15.

x=7

Divide 14 by 2.

x=\frac{-16}{2}

Now solve the equation x=\frac{-1±15}{2} when ± is minus. Subtract 15 from -1.

x=-8

Divide -16 by 2.

x=7 x=-8

The equation is now solved.

x^{2}+x-56=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}+x-56-\left(-56\right)=-\left(-56\right)

Add 56 to both sides of the equation.

x^{2}+x=-\left(-56\right)

Subtracting -56 from itself leaves 0.

x^{2}+x=56

Subtract -56 from 0.

x^{2}+x+\left(\frac{1}{2}\right)^{2}=56+\left(\frac{1}{2}\right)^{2}

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}+x+\frac{1}{4}=56+\frac{1}{4}

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

x^{2}+x+\frac{1}{4}=\frac{225}{4}

Add 56 to \frac{1}{4}.

\left(x+\frac{1}{2}\right)^{2}=\frac{225}{4}

Factor x^{2}+x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{225}{4}}

Take the square root of both sides of the equation.

x+\frac{1}{2}=\frac{15}{2} x+\frac{1}{2}=-\frac{15}{2}

Simplify.

x=7 x=-8

Subtract \frac{1}{2} from both sides of the equation.

x ^ 2 +1x -56 = 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 = -1 rs = -56

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

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>

(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -56

To solve for unknown quantity u, substitute these in the product equation rs = -56

\frac{1}{4} - u^2 = -56

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

-u^2 = -56-\frac{1}{4} = -\frac{225}{4}

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

u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{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{1}{2} - \frac{15}{2} = -8 s = -\frac{1}{2} + \frac{15}{2} = 7

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

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