Solve for a
a=2\sqrt{5}-4\approx 0.472135955
a=-2\sqrt{5}-4\approx -8.472135955
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a^{2}+8a-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.
a=\frac{-8±\sqrt{8^{2}-4\left(-4\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-8±\sqrt{64-4\left(-4\right)}}{2}
Square 8.
a=\frac{-8±\sqrt{64+16}}{2}
Multiply -4 times -4.
a=\frac{-8±\sqrt{80}}{2}
Add 64 to 16.
a=\frac{-8±4\sqrt{5}}{2}
Take the square root of 80.
a=\frac{4\sqrt{5}-8}{2}
Now solve the equation a=\frac{-8±4\sqrt{5}}{2} when ± is plus. Add -8 to 4\sqrt{5}.
a=2\sqrt{5}-4
Divide -8+4\sqrt{5} by 2.
a=\frac{-4\sqrt{5}-8}{2}
Now solve the equation a=\frac{-8±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from -8.
a=-2\sqrt{5}-4
Divide -8-4\sqrt{5} by 2.
a=2\sqrt{5}-4 a=-2\sqrt{5}-4
The equation is now solved.
a^{2}+8a-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.
a^{2}+8a-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
a^{2}+8a=-\left(-4\right)
Subtracting -4 from itself leaves 0.
a^{2}+8a=4
Subtract -4 from 0.
a^{2}+8a+4^{2}=4+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}+8a+16=4+16
Square 4.
a^{2}+8a+16=20
Add 4 to 16.
\left(a+4\right)^{2}=20
Factor a^{2}+8a+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+4\right)^{2}}=\sqrt{20}
Take the square root of both sides of the equation.
a+4=2\sqrt{5} a+4=-2\sqrt{5}
Simplify.
a=2\sqrt{5}-4 a=-2\sqrt{5}-4
Subtract 4 from both sides of the equation.
x ^ 2 +8x -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.
r + s = -8 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 = -4 - u s = -4 + u
Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -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>
(-4 - u) (-4 + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
16 - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-16 = -20
Simplify the expression by subtracting 16 on both sides
u^2 = 20 u = \pm\sqrt{20} = \pm \sqrt{20}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-4 - \sqrt{20} = -8.472 s = -4 + \sqrt{20} = 0.472
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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Simultaneous equation
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Limits
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