Solve for a
a = \frac{\sqrt{113} + 7}{4} \approx 4.407536453
a=\frac{7-\sqrt{113}}{4}\approx -0.907536453
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2a^{2}-7a-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.
a=\frac{-\left(-7\right)±\sqrt{\left(-7\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, -7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-8\right)}}{2\times 2}
Square -7.
a=\frac{-\left(-7\right)±\sqrt{49-8\left(-8\right)}}{2\times 2}
Multiply -4 times 2.
a=\frac{-\left(-7\right)±\sqrt{49+64}}{2\times 2}
Multiply -8 times -8.
a=\frac{-\left(-7\right)±\sqrt{113}}{2\times 2}
Add 49 to 64.
a=\frac{7±\sqrt{113}}{2\times 2}
The opposite of -7 is 7.
a=\frac{7±\sqrt{113}}{4}
Multiply 2 times 2.
a=\frac{\sqrt{113}+7}{4}
Now solve the equation a=\frac{7±\sqrt{113}}{4} when ± is plus. Add 7 to \sqrt{113}.
a=\frac{7-\sqrt{113}}{4}
Now solve the equation a=\frac{7±\sqrt{113}}{4} when ± is minus. Subtract \sqrt{113} from 7.
a=\frac{\sqrt{113}+7}{4} a=\frac{7-\sqrt{113}}{4}
The equation is now solved.
2a^{2}-7a-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.
2a^{2}-7a-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
2a^{2}-7a=-\left(-8\right)
Subtracting -8 from itself leaves 0.
2a^{2}-7a=8
Subtract -8 from 0.
\frac{2a^{2}-7a}{2}=\frac{8}{2}
Divide both sides by 2.
a^{2}-\frac{7}{2}a=\frac{8}{2}
Dividing by 2 undoes the multiplication by 2.
a^{2}-\frac{7}{2}a=4
Divide 8 by 2.
a^{2}-\frac{7}{2}a+\left(-\frac{7}{4}\right)^{2}=4+\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.
a^{2}-\frac{7}{2}a+\frac{49}{16}=4+\frac{49}{16}
Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{7}{2}a+\frac{49}{16}=\frac{113}{16}
Add 4 to \frac{49}{16}.
\left(a-\frac{7}{4}\right)^{2}=\frac{113}{16}
Factor a^{2}-\frac{7}{2}a+\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(a-\frac{7}{4}\right)^{2}}=\sqrt{\frac{113}{16}}
Take the square root of both sides of the equation.
a-\frac{7}{4}=\frac{\sqrt{113}}{4} a-\frac{7}{4}=-\frac{\sqrt{113}}{4}
Simplify.
a=\frac{\sqrt{113}+7}{4} a=\frac{7-\sqrt{113}}{4}
Add \frac{7}{4} to both sides of the equation.
x ^ 2 -\frac{7}{2}x -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 = \frac{7}{2} 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 = \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) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
\frac{49}{16} - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-\frac{49}{16} = -\frac{113}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{113}{16} u = \pm\sqrt{\frac{113}{16}} = \pm \frac{\sqrt{113}}{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{113}}{4} = -0.908 s = \frac{7}{4} + \frac{\sqrt{113}}{4} = 4.408
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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