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x^{2}+62x+1=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{-62±\sqrt{62^{2}-4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 62 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-62±\sqrt{3844-4}}{2}
Square 62.
x=\frac{-62±\sqrt{3840}}{2}
Add 3844 to -4.
x=\frac{-62±16\sqrt{15}}{2}
Take the square root of 3840.
x=\frac{16\sqrt{15}-62}{2}
Now solve the equation x=\frac{-62±16\sqrt{15}}{2} when ± is plus. Add -62 to 16\sqrt{15}.
x=8\sqrt{15}-31
Divide -62+16\sqrt{15} by 2.
x=\frac{-16\sqrt{15}-62}{2}
Now solve the equation x=\frac{-62±16\sqrt{15}}{2} when ± is minus. Subtract 16\sqrt{15} from -62.
x=-8\sqrt{15}-31
Divide -62-16\sqrt{15} by 2.
x=8\sqrt{15}-31 x=-8\sqrt{15}-31
The equation is now solved.
x^{2}+62x+1=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}+62x+1-1=-1
Subtract 1 from both sides of the equation.
x^{2}+62x=-1
Subtracting 1 from itself leaves 0.
x^{2}+62x+31^{2}=-1+31^{2}
Divide 62, the coefficient of the x term, by 2 to get 31. Then add the square of 31 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+62x+961=-1+961
Square 31.
x^{2}+62x+961=960
Add -1 to 961.
\left(x+31\right)^{2}=960
Factor x^{2}+62x+961. 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+31\right)^{2}}=\sqrt{960}
Take the square root of both sides of the equation.
x+31=8\sqrt{15} x+31=-8\sqrt{15}
Simplify.
x=8\sqrt{15}-31 x=-8\sqrt{15}-31
Subtract 31 from both sides of the equation.
x ^ 2 +62x +1 = 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 = -62 rs = 1
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 = -31 - u s = -31 + u
Two numbers r and s sum up to -62 exactly when the average of the two numbers is \frac{1}{2}*-62 = -31. 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>
(-31 - u) (-31 + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
961 - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-961 = -960
Simplify the expression by subtracting 961 on both sides
u^2 = 960 u = \pm\sqrt{960} = \pm \sqrt{960}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-31 - \sqrt{960} = -61.984 s = -31 + \sqrt{960} = -0.016
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.