Solve for x
x = \frac{\sqrt{129} - 9}{2} \approx 1.178908346
x=\frac{-\sqrt{129}-9}{2}\approx -10.178908346
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x^{2}+9x-12=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{-9±\sqrt{9^{2}-4\left(-12\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-9±\sqrt{81-4\left(-12\right)}}{2}
Square 9.
x=\frac{-9±\sqrt{81+48}}{2}
Multiply -4 times -12.
x=\frac{-9±\sqrt{129}}{2}
Add 81 to 48.
x=\frac{\sqrt{129}-9}{2}
Now solve the equation x=\frac{-9±\sqrt{129}}{2} when ± is plus. Add -9 to \sqrt{129}.
x=\frac{-\sqrt{129}-9}{2}
Now solve the equation x=\frac{-9±\sqrt{129}}{2} when ± is minus. Subtract \sqrt{129} from -9.
x=\frac{\sqrt{129}-9}{2} x=\frac{-\sqrt{129}-9}{2}
The equation is now solved.
x^{2}+9x-12=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.
x^{2}+9x-12-\left(-12\right)=-\left(-12\right)
Add 12 to both sides of the equation.
x^{2}+9x=-\left(-12\right)
Subtracting -12 from itself leaves 0.
x^{2}+9x=12
Subtract -12 from 0.
x^{2}+9x+\left(\frac{9}{2}\right)^{2}=12+\left(\frac{9}{2}\right)^{2}
Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+9x+\frac{81}{4}=12+\frac{81}{4}
Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+9x+\frac{81}{4}=\frac{129}{4}
Add 12 to \frac{81}{4}.
\left(x+\frac{9}{2}\right)^{2}=\frac{129}{4}
Factor x^{2}+9x+\frac{81}{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+\frac{9}{2}\right)^{2}}=\sqrt{\frac{129}{4}}
Take the square root of both sides of the equation.
x+\frac{9}{2}=\frac{\sqrt{129}}{2} x+\frac{9}{2}=-\frac{\sqrt{129}}{2}
Simplify.
x=\frac{\sqrt{129}-9}{2} x=\frac{-\sqrt{129}-9}{2}
Subtract \frac{9}{2} from both sides of the equation.
x ^ 2 +9x -12 = 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 = -9 rs = -12
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{9}{2} - u s = -\frac{9}{2} + u
Two numbers r and s sum up to -9 exactly when the average of the two numbers is \frac{1}{2}*-9 = -\frac{9}{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>
(-\frac{9}{2} - u) (-\frac{9}{2} + u) = -12
To solve for unknown quantity u, substitute these in the product equation rs = -12
\frac{81}{4} - u^2 = -12
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -12-\frac{81}{4} = -\frac{129}{4}
Simplify the expression by subtracting \frac{81}{4} on both sides
u^2 = \frac{129}{4} u = \pm\sqrt{\frac{129}{4}} = \pm \frac{\sqrt{129}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{9}{2} - \frac{\sqrt{129}}{2} = -10.179 s = -\frac{9}{2} + \frac{\sqrt{129}}{2} = 1.179
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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Differentiation
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Limits
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