Solve for x
x=\frac{1}{8}=0.125
x=2
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8x^{2}-17x+2=0
Divide both sides by 4.
a+b=-17 ab=8\times 2=16
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
-1,-16 -2,-8 -4,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.
-1-16=-17 -2-8=-10 -4-4=-8
Calculate the sum for each pair.
a=-16 b=-1
The solution is the pair that gives sum -17.
\left(8x^{2}-16x\right)+\left(-x+2\right)
Rewrite 8x^{2}-17x+2 as \left(8x^{2}-16x\right)+\left(-x+2\right).
8x\left(x-2\right)-\left(x-2\right)
Factor out 8x in the first and -1 in the second group.
\left(x-2\right)\left(8x-1\right)
Factor out common term x-2 by using distributive property.
x=2 x=\frac{1}{8}
To find equation solutions, solve x-2=0 and 8x-1=0.
32x^{2}-68x+8=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(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 32\times 8}}{2\times 32}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 32 for a, -68 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-68\right)±\sqrt{4624-4\times 32\times 8}}{2\times 32}
Square -68.
x=\frac{-\left(-68\right)±\sqrt{4624-128\times 8}}{2\times 32}
Multiply -4 times 32.
x=\frac{-\left(-68\right)±\sqrt{4624-1024}}{2\times 32}
Multiply -128 times 8.
x=\frac{-\left(-68\right)±\sqrt{3600}}{2\times 32}
Add 4624 to -1024.
x=\frac{-\left(-68\right)±60}{2\times 32}
Take the square root of 3600.
x=\frac{68±60}{2\times 32}
The opposite of -68 is 68.
x=\frac{68±60}{64}
Multiply 2 times 32.
x=\frac{128}{64}
Now solve the equation x=\frac{68±60}{64} when ± is plus. Add 68 to 60.
x=2
Divide 128 by 64.
x=\frac{8}{64}
Now solve the equation x=\frac{68±60}{64} when ± is minus. Subtract 60 from 68.
x=\frac{1}{8}
Reduce the fraction \frac{8}{64} to lowest terms by extracting and canceling out 8.
x=2 x=\frac{1}{8}
The equation is now solved.
32x^{2}-68x+8=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.
32x^{2}-68x+8-8=-8
Subtract 8 from both sides of the equation.
32x^{2}-68x=-8
Subtracting 8 from itself leaves 0.
\frac{32x^{2}-68x}{32}=-\frac{8}{32}
Divide both sides by 32.
x^{2}+\left(-\frac{68}{32}\right)x=-\frac{8}{32}
Dividing by 32 undoes the multiplication by 32.
x^{2}-\frac{17}{8}x=-\frac{8}{32}
Reduce the fraction \frac{-68}{32} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{17}{8}x=-\frac{1}{4}
Reduce the fraction \frac{-8}{32} to lowest terms by extracting and canceling out 8.
x^{2}-\frac{17}{8}x+\left(-\frac{17}{16}\right)^{2}=-\frac{1}{4}+\left(-\frac{17}{16}\right)^{2}
Divide -\frac{17}{8}, the coefficient of the x term, by 2 to get -\frac{17}{16}. Then add the square of -\frac{17}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{17}{8}x+\frac{289}{256}=-\frac{1}{4}+\frac{289}{256}
Square -\frac{17}{16} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{17}{8}x+\frac{289}{256}=\frac{225}{256}
Add -\frac{1}{4} to \frac{289}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{17}{16}\right)^{2}=\frac{225}{256}
Factor x^{2}-\frac{17}{8}x+\frac{289}{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{17}{16}\right)^{2}}=\sqrt{\frac{225}{256}}
Take the square root of both sides of the equation.
x-\frac{17}{16}=\frac{15}{16} x-\frac{17}{16}=-\frac{15}{16}
Simplify.
x=2 x=\frac{1}{8}
Add \frac{17}{16} to both sides of the equation.
x ^ 2 -\frac{17}{8}x +\frac{1}{4} = 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.This is achieved by dividing both sides of the equation by 32
r + s = \frac{17}{8} rs = \frac{1}{4}
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{17}{16} - u s = \frac{17}{16} + u
Two numbers r and s sum up to \frac{17}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{8} = \frac{17}{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{17}{16} - u) (\frac{17}{16} + u) = \frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}
\frac{289}{256} - u^2 = \frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{4}-\frac{289}{256} = -\frac{225}{256}
Simplify the expression by subtracting \frac{289}{256} on both sides
u^2 = \frac{225}{256} u = \pm\sqrt{\frac{225}{256}} = \pm \frac{15}{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{17}{16} - \frac{15}{16} = 0.125 s = \frac{17}{16} + \frac{15}{16} = 2
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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