Solve for y
y=\frac{25+\sqrt{399}i}{2}\approx 12.5+9.987492178i
y=\frac{-\sqrt{399}i+25}{2}\approx 12.5-9.987492178i
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y^{2}-25y+256=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.
y=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 256}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -25 for b, and 256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-25\right)±\sqrt{625-4\times 256}}{2}
Square -25.
y=\frac{-\left(-25\right)±\sqrt{625-1024}}{2}
Multiply -4 times 256.
y=\frac{-\left(-25\right)±\sqrt{-399}}{2}
Add 625 to -1024.
y=\frac{-\left(-25\right)±\sqrt{399}i}{2}
Take the square root of -399.
y=\frac{25±\sqrt{399}i}{2}
The opposite of -25 is 25.
y=\frac{25+\sqrt{399}i}{2}
Now solve the equation y=\frac{25±\sqrt{399}i}{2} when ± is plus. Add 25 to i\sqrt{399}.
y=\frac{-\sqrt{399}i+25}{2}
Now solve the equation y=\frac{25±\sqrt{399}i}{2} when ± is minus. Subtract i\sqrt{399} from 25.
y=\frac{25+\sqrt{399}i}{2} y=\frac{-\sqrt{399}i+25}{2}
The equation is now solved.
y^{2}-25y+256=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.
y^{2}-25y+256-256=-256
Subtract 256 from both sides of the equation.
y^{2}-25y=-256
Subtracting 256 from itself leaves 0.
y^{2}-25y+\left(-\frac{25}{2}\right)^{2}=-256+\left(-\frac{25}{2}\right)^{2}
Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-25y+\frac{625}{4}=-256+\frac{625}{4}
Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.
y^{2}-25y+\frac{625}{4}=-\frac{399}{4}
Add -256 to \frac{625}{4}.
\left(y-\frac{25}{2}\right)^{2}=-\frac{399}{4}
Factor y^{2}-25y+\frac{625}{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(y-\frac{25}{2}\right)^{2}}=\sqrt{-\frac{399}{4}}
Take the square root of both sides of the equation.
y-\frac{25}{2}=\frac{\sqrt{399}i}{2} y-\frac{25}{2}=-\frac{\sqrt{399}i}{2}
Simplify.
y=\frac{25+\sqrt{399}i}{2} y=\frac{-\sqrt{399}i+25}{2}
Add \frac{25}{2} to both sides of the equation.
x ^ 2 -25x +256 = 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 = 25 rs = 256
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{25}{2} - u s = \frac{25}{2} + u
Two numbers r and s sum up to 25 exactly when the average of the two numbers is \frac{1}{2}*25 = \frac{25}{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{25}{2} - u) (\frac{25}{2} + u) = 256
To solve for unknown quantity u, substitute these in the product equation rs = 256
\frac{625}{4} - u^2 = 256
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 256-\frac{625}{4} = \frac{399}{4}
Simplify the expression by subtracting \frac{625}{4} on both sides
u^2 = -\frac{399}{4} u = \pm\sqrt{-\frac{399}{4}} = \pm \frac{\sqrt{399}}{2}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{25}{2} - \frac{\sqrt{399}}{2}i = 12.500 - 9.987i s = \frac{25}{2} + \frac{\sqrt{399}}{2}i = 12.500 + 9.987i
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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