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16x^{2}+64x+65=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{-64±\sqrt{64^{2}-4\times 16\times 65}}{2\times 16}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 64 for b, and 65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-64±\sqrt{4096-4\times 16\times 65}}{2\times 16}
Square 64.
x=\frac{-64±\sqrt{4096-64\times 65}}{2\times 16}
Multiply -4 times 16.
x=\frac{-64±\sqrt{4096-4160}}{2\times 16}
Multiply -64 times 65.
x=\frac{-64±\sqrt{-64}}{2\times 16}
Add 4096 to -4160.
x=\frac{-64±8i}{2\times 16}
Take the square root of -64.
x=\frac{-64±8i}{32}
Multiply 2 times 16.
x=\frac{-64+8i}{32}
Now solve the equation x=\frac{-64±8i}{32} when ± is plus. Add -64 to 8i.
x=-2+\frac{1}{4}i
Divide -64+8i by 32.
x=\frac{-64-8i}{32}
Now solve the equation x=\frac{-64±8i}{32} when ± is minus. Subtract 8i from -64.
x=-2-\frac{1}{4}i
Divide -64-8i by 32.
x=-2+\frac{1}{4}i x=-2-\frac{1}{4}i
The equation is now solved.
16x^{2}+64x+65=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.
16x^{2}+64x+65-65=-65
Subtract 65 from both sides of the equation.
16x^{2}+64x=-65
Subtracting 65 from itself leaves 0.
\frac{16x^{2}+64x}{16}=-\frac{65}{16}
Divide both sides by 16.
x^{2}+\frac{64}{16}x=-\frac{65}{16}
Dividing by 16 undoes the multiplication by 16.
x^{2}+4x=-\frac{65}{16}
Divide 64 by 16.
x^{2}+4x+2^{2}=-\frac{65}{16}+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=-\frac{65}{16}+4
Square 2.
x^{2}+4x+4=-\frac{1}{16}
Add -\frac{65}{16} to 4.
\left(x+2\right)^{2}=-\frac{1}{16}
Factor x^{2}+4x+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+2\right)^{2}}=\sqrt{-\frac{1}{16}}
Take the square root of both sides of the equation.
x+2=\frac{1}{4}i x+2=-\frac{1}{4}i
Simplify.
x=-2+\frac{1}{4}i x=-2-\frac{1}{4}i
Subtract 2 from both sides of the equation.
x ^ 2 +4x +\frac{65}{16} = 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 16
r + s = -4 rs = \frac{65}{16}
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 = -2 - u s = -2 + u
Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>
(-2 - u) (-2 + u) = \frac{65}{16}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{65}{16}
4 - u^2 = \frac{65}{16}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{65}{16}-4 = \frac{1}{16}
Simplify the expression by subtracting 4 on both sides
u^2 = -\frac{1}{16} u = \pm\sqrt{-\frac{1}{16}} = \pm \frac{1}{4}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-2 - \frac{1}{4}i = -2 - 0.250i s = -2 + \frac{1}{4}i = -2 + 0.250i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.