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-2x^{2}+7x-2=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-7±\sqrt{7^{2}-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}
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{-7±\sqrt{49-4\left(-2\right)\left(-2\right)}}{2\left(-2\right)}
Square 7.
x=\frac{-7±\sqrt{49+8\left(-2\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-7±\sqrt{49-16}}{2\left(-2\right)}
Multiply 8 times -2.
x=\frac{-7±\sqrt{33}}{2\left(-2\right)}
Add 49 to -16.
x=\frac{-7±\sqrt{33}}{-4}
Multiply 2 times -2.
x=\frac{\sqrt{33}-7}{-4}
Now solve the equation x=\frac{-7±\sqrt{33}}{-4} when ± is plus. Add -7 to \sqrt{33}.
x=\frac{7-\sqrt{33}}{4}
Divide -7+\sqrt{33} by -4.
x=\frac{-\sqrt{33}-7}{-4}
Now solve the equation x=\frac{-7±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from -7.
x=\frac{\sqrt{33}+7}{4}
Divide -7-\sqrt{33} by -4.
-2x^{2}+7x-2=-2\left(x-\frac{7-\sqrt{33}}{4}\right)\left(x-\frac{\sqrt{33}+7}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7-\sqrt{33}}{4} for x_{1} and \frac{7+\sqrt{33}}{4} for x_{2}.
x ^ 2 -\frac{7}{2}x +1 = 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}{2} rs = 1
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}{4} - u s = \frac{7}{4} + u
Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{4}. 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}{4} - u) (\frac{7}{4} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{49}{16} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{49}{16} = -\frac{33}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{33}{16} u = \pm\sqrt{\frac{33}{16}} = \pm \frac{\sqrt{33}}{4}
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}{4} - \frac{\sqrt{33}}{4} = 0.314 s = \frac{7}{4} + \frac{\sqrt{33}}{4} = 3.186
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.