Solve for x
x=\frac{1}{8}=0.125
x=-1
Graph
Share
Copied to clipboard
a+b=-7 ab=-8=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.
1,-8 2,-4
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 -8.
1-8=-7 2-4=-2
Calculate the sum for each pair.
a=1 b=-8
The solution is the pair that gives sum -7.
\left(-8x^{2}+x\right)+\left(-8x+1\right)
Rewrite -8x^{2}-7x+1 as \left(-8x^{2}+x\right)+\left(-8x+1\right).
-x\left(8x-1\right)-\left(8x-1\right)
Factor out -x in the first and -1 in the second group.
\left(8x-1\right)\left(-x-1\right)
Factor out common term 8x-1 by using distributive property.
x=\frac{1}{8} x=-1
To find equation solutions, solve 8x-1=0 and -x-1=0.
-8x^{2}-7x+1=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-8\right)}}{2\left(-8\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, -7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\left(-8\right)}}{2\left(-8\right)}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\left(-8\right)}
Multiply -4 times -8.
x=\frac{-\left(-7\right)±\sqrt{81}}{2\left(-8\right)}
Add 49 to 32.
x=\frac{-\left(-7\right)±9}{2\left(-8\right)}
Take the square root of 81.
x=\frac{7±9}{2\left(-8\right)}
The opposite of -7 is 7.
x=\frac{7±9}{-16}
Multiply 2 times -8.
x=\frac{16}{-16}
Now solve the equation x=\frac{7±9}{-16} when ± is plus. Add 7 to 9.
x=-1
Divide 16 by -16.
x=-\frac{2}{-16}
Now solve the equation x=\frac{7±9}{-16} when ± is minus. Subtract 9 from 7.
x=\frac{1}{8}
Reduce the fraction \frac{-2}{-16} to lowest terms by extracting and canceling out 2.
x=-1 x=\frac{1}{8}
The equation is now solved.
-8x^{2}-7x+1=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.
-8x^{2}-7x+1-1=-1
Subtract 1 from both sides of the equation.
-8x^{2}-7x=-1
Subtracting 1 from itself leaves 0.
\frac{-8x^{2}-7x}{-8}=-\frac{1}{-8}
Divide both sides by -8.
x^{2}+\left(-\frac{7}{-8}\right)x=-\frac{1}{-8}
Dividing by -8 undoes the multiplication by -8.
x^{2}+\frac{7}{8}x=-\frac{1}{-8}
Divide -7 by -8.
x^{2}+\frac{7}{8}x=\frac{1}{8}
Divide -1 by -8.
x^{2}+\frac{7}{8}x+\left(\frac{7}{16}\right)^{2}=\frac{1}{8}+\left(\frac{7}{16}\right)^{2}
Divide \frac{7}{8}, the coefficient of the x term, by 2 to get \frac{7}{16}. Then add the square of \frac{7}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{7}{8}x+\frac{49}{256}=\frac{1}{8}+\frac{49}{256}
Square \frac{7}{16} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{8}x+\frac{49}{256}=\frac{81}{256}
Add \frac{1}{8} to \frac{49}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{7}{16}\right)^{2}=\frac{81}{256}
Factor x^{2}+\frac{7}{8}x+\frac{49}{256}. 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{7}{16}\right)^{2}}=\sqrt{\frac{81}{256}}
Take the square root of both sides of the equation.
x+\frac{7}{16}=\frac{9}{16} x+\frac{7}{16}=-\frac{9}{16}
Simplify.
x=\frac{1}{8} x=-1
Subtract \frac{7}{16} from both sides of the equation.
x ^ 2 +\frac{7}{8}x -\frac{1}{8} = 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 = -\frac{7}{8} rs = -\frac{1}{8}
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{7}{16} - u s = -\frac{7}{16} + u
Two numbers r and s sum up to -\frac{7}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{8} = -\frac{7}{16}. 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{7}{16} - u) (-\frac{7}{16} + u) = -\frac{1}{8}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{8}
\frac{49}{256} - u^2 = -\frac{1}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{8}-\frac{49}{256} = -\frac{81}{256}
Simplify the expression by subtracting \frac{49}{256} on both sides
u^2 = \frac{81}{256} u = \pm\sqrt{\frac{81}{256}} = \pm \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
r =-\frac{7}{16} - \frac{9}{16} = -1 s = -\frac{7}{16} + \frac{9}{16} = 0.125
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}