Solve for k
k=\frac{\sqrt{10}}{4}+\frac{1}{2}\approx 1.290569415
k=-\frac{\sqrt{10}}{4}+\frac{1}{2}\approx -0.290569415
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-24k^{2}+24k+9=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.
k=\frac{-24±\sqrt{24^{2}-4\left(-24\right)\times 9}}{2\left(-24\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -24 for a, 24 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-24±\sqrt{576-4\left(-24\right)\times 9}}{2\left(-24\right)}
Square 24.
k=\frac{-24±\sqrt{576+96\times 9}}{2\left(-24\right)}
Multiply -4 times -24.
k=\frac{-24±\sqrt{576+864}}{2\left(-24\right)}
Multiply 96 times 9.
k=\frac{-24±\sqrt{1440}}{2\left(-24\right)}
Add 576 to 864.
k=\frac{-24±12\sqrt{10}}{2\left(-24\right)}
Take the square root of 1440.
k=\frac{-24±12\sqrt{10}}{-48}
Multiply 2 times -24.
k=\frac{12\sqrt{10}-24}{-48}
Now solve the equation k=\frac{-24±12\sqrt{10}}{-48} when ± is plus. Add -24 to 12\sqrt{10}.
k=-\frac{\sqrt{10}}{4}+\frac{1}{2}
Divide -24+12\sqrt{10} by -48.
k=\frac{-12\sqrt{10}-24}{-48}
Now solve the equation k=\frac{-24±12\sqrt{10}}{-48} when ± is minus. Subtract 12\sqrt{10} from -24.
k=\frac{\sqrt{10}}{4}+\frac{1}{2}
Divide -24-12\sqrt{10} by -48.
k=-\frac{\sqrt{10}}{4}+\frac{1}{2} k=\frac{\sqrt{10}}{4}+\frac{1}{2}
The equation is now solved.
-24k^{2}+24k+9=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.
-24k^{2}+24k+9-9=-9
Subtract 9 from both sides of the equation.
-24k^{2}+24k=-9
Subtracting 9 from itself leaves 0.
\frac{-24k^{2}+24k}{-24}=-\frac{9}{-24}
Divide both sides by -24.
k^{2}+\frac{24}{-24}k=-\frac{9}{-24}
Dividing by -24 undoes the multiplication by -24.
k^{2}-k=-\frac{9}{-24}
Divide 24 by -24.
k^{2}-k=\frac{3}{8}
Reduce the fraction \frac{-9}{-24} to lowest terms by extracting and canceling out 3.
k^{2}-k+\left(-\frac{1}{2}\right)^{2}=\frac{3}{8}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-k+\frac{1}{4}=\frac{3}{8}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
k^{2}-k+\frac{1}{4}=\frac{5}{8}
Add \frac{3}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k-\frac{1}{2}\right)^{2}=\frac{5}{8}
Factor k^{2}-k+\frac{1}{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(k-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{8}}
Take the square root of both sides of the equation.
k-\frac{1}{2}=\frac{\sqrt{10}}{4} k-\frac{1}{2}=-\frac{\sqrt{10}}{4}
Simplify.
k=\frac{\sqrt{10}}{4}+\frac{1}{2} k=-\frac{\sqrt{10}}{4}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -\frac{3}{8} = 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 = 1 rs = -\frac{3}{8}
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{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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{1}{2} - u) (\frac{1}{2} + u) = -\frac{3}{8}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{8}
\frac{1}{4} - u^2 = -\frac{3}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{8}-\frac{1}{4} = -\frac{5}{8}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{5}{8} u = \pm\sqrt{\frac{5}{8}} = \pm \frac{\sqrt{5}}{\sqrt{8}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - \frac{\sqrt{5}}{\sqrt{8}} = -0.291 s = \frac{1}{2} + \frac{\sqrt{5}}{\sqrt{8}} = 1.291
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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