Solve for x
x=-4
x=\frac{1}{2}=0.5
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a+b=7 ab=2\left(-4\right)=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-4. 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 positive, the positive number has greater absolute value than the negative. 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(2x^{2}-x\right)+\left(8x-4\right)
Rewrite 2x^{2}+7x-4 as \left(2x^{2}-x\right)+\left(8x-4\right).
x\left(2x-1\right)+4\left(2x-1\right)
Factor out x in the first and 4 in the second group.
\left(2x-1\right)\left(x+4\right)
Factor out common term 2x-1 by using distributive property.
x=\frac{1}{2} x=-4
To find equation solutions, solve 2x-1=0 and x+4=0.
2x^{2}+7x-4=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{-7±\sqrt{7^{2}-4\times 2\left(-4\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 2\left(-4\right)}}{2\times 2}
Square 7.
x=\frac{-7±\sqrt{49-8\left(-4\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-7±\sqrt{49+32}}{2\times 2}
Multiply -8 times -4.
x=\frac{-7±\sqrt{81}}{2\times 2}
Add 49 to 32.
x=\frac{-7±9}{2\times 2}
Take the square root of 81.
x=\frac{-7±9}{4}
Multiply 2 times 2.
x=\frac{2}{4}
Now solve the equation x=\frac{-7±9}{4} when ± is plus. Add -7 to 9.
x=\frac{1}{2}
Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.
x=-\frac{16}{4}
Now solve the equation x=\frac{-7±9}{4} when ± is minus. Subtract 9 from -7.
x=-4
Divide -16 by 4.
x=\frac{1}{2} x=-4
The equation is now solved.
2x^{2}+7x-4=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}+7x-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
2x^{2}+7x=-\left(-4\right)
Subtracting -4 from itself leaves 0.
2x^{2}+7x=4
Subtract -4 from 0.
\frac{2x^{2}+7x}{2}=\frac{4}{2}
Divide both sides by 2.
x^{2}+\frac{7}{2}x=\frac{4}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{7}{2}x=2
Divide 4 by 2.
x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=2+\left(\frac{7}{4}\right)^{2}
Divide \frac{7}{2}, the coefficient of the x term, by 2 to get \frac{7}{4}. Then add the square of \frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{7}{2}x+\frac{49}{16}=2+\frac{49}{16}
Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{81}{16}
Add 2 to \frac{49}{16}.
\left(x+\frac{7}{4}\right)^{2}=\frac{81}{16}
Factor x^{2}+\frac{7}{2}x+\frac{49}{16}. 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}{4}\right)^{2}}=\sqrt{\frac{81}{16}}
Take the square root of both sides of the equation.
x+\frac{7}{4}=\frac{9}{4} x+\frac{7}{4}=-\frac{9}{4}
Simplify.
x=\frac{1}{2} x=-4
Subtract \frac{7}{4} from both sides of the equation.
x ^ 2 +\frac{7}{2}x -2 = 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 = -\frac{7}{2} rs = -2
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) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{49}{16} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{49}{16} = -\frac{81}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{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{9}{4} = -4 s = -\frac{7}{4} + \frac{9}{4} = 0.500
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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Limits
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