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2x^{2}-8x-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(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-8\right)}}{2\times 2}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64-8\left(-8\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-8\right)±\sqrt{64+64}}{2\times 2}
Multiply -8 times -8.
x=\frac{-\left(-8\right)±\sqrt{128}}{2\times 2}
Add 64 to 64.
x=\frac{-\left(-8\right)±8\sqrt{2}}{2\times 2}
Take the square root of 128.
x=\frac{8±8\sqrt{2}}{2\times 2}
The opposite of -8 is 8.
x=\frac{8±8\sqrt{2}}{4}
Multiply 2 times 2.
x=\frac{8\sqrt{2}+8}{4}
Now solve the equation x=\frac{8±8\sqrt{2}}{4} when ± is plus. Add 8 to 8\sqrt{2}.
x=2\sqrt{2}+2
Divide 8+8\sqrt{2} by 4.
x=\frac{8-8\sqrt{2}}{4}
Now solve the equation x=\frac{8±8\sqrt{2}}{4} when ± is minus. Subtract 8\sqrt{2} from 8.
x=2-2\sqrt{2}
Divide 8-8\sqrt{2} by 4.
x=2\sqrt{2}+2 x=2-2\sqrt{2}
The equation is now solved.
2x^{2}-8x-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.
2x^{2}-8x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
2x^{2}-8x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
2x^{2}-8x=8
Subtract -8 from 0.
\frac{2x^{2}-8x}{2}=\frac{8}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{8}{2}\right)x=\frac{8}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-4x=\frac{8}{2}
Divide -8 by 2.
x^{2}-4x=4
Divide 8 by 2.
x^{2}-4x+\left(-2\right)^{2}=4+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=4+4
Square -2.
x^{2}-4x+4=8
Add 4 to 4.
\left(x-2\right)^{2}=8
Factor x^{2}-4x+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-2\right)^{2}}=\sqrt{8}
Take the square root of both sides of the equation.
x-2=2\sqrt{2} x-2=-2\sqrt{2}
Simplify.
x=2\sqrt{2}+2 x=2-2\sqrt{2}
Add 2 to both sides of the equation.
x ^ 2 -4x -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 2
r + s = 4 rs = -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 = 2 - u s = 2 + u
Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 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>
(2 - u) (2 + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
4 - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-4 = -8
Simplify the expression by subtracting 4 on both sides
u^2 = 8 u = \pm\sqrt{8} = \pm \sqrt{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =2 - \sqrt{8} = -0.828 s = 2 + \sqrt{8} = 4.828
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.