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w^{2}-10w+5=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.
w=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
w=\frac{-\left(-10\right)±\sqrt{100-4\times 5}}{2}
Square -10.
w=\frac{-\left(-10\right)±\sqrt{100-20}}{2}
Multiply -4 times 5.
w=\frac{-\left(-10\right)±\sqrt{80}}{2}
Add 100 to -20.
w=\frac{-\left(-10\right)±4\sqrt{5}}{2}
Take the square root of 80.
w=\frac{10±4\sqrt{5}}{2}
The opposite of -10 is 10.
w=\frac{4\sqrt{5}+10}{2}
Now solve the equation w=\frac{10±4\sqrt{5}}{2} when ± is plus. Add 10 to 4\sqrt{5}.
w=2\sqrt{5}+5
Divide 10+4\sqrt{5} by 2.
w=\frac{10-4\sqrt{5}}{2}
Now solve the equation w=\frac{10±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 10.
w=5-2\sqrt{5}
Divide 10-4\sqrt{5} by 2.
w=2\sqrt{5}+5 w=5-2\sqrt{5}
The equation is now solved.
w^{2}-10w+5=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.
w^{2}-10w+5-5=-5
Subtract 5 from both sides of the equation.
w^{2}-10w=-5
Subtracting 5 from itself leaves 0.
w^{2}-10w+\left(-5\right)^{2}=-5+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
w^{2}-10w+25=-5+25
Square -5.
w^{2}-10w+25=20
Add -5 to 25.
\left(w-5\right)^{2}=20
Factor w^{2}-10w+25. 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(w-5\right)^{2}}=\sqrt{20}
Take the square root of both sides of the equation.
w-5=2\sqrt{5} w-5=-2\sqrt{5}
Simplify.
w=2\sqrt{5}+5 w=5-2\sqrt{5}
Add 5 to both sides of the equation.
x ^ 2 -10x +5 = 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 = 10 rs = 5
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 = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>
(5 - u) (5 + u) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
25 - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-25 = -20
Simplify the expression by subtracting 25 on both sides
u^2 = 20 u = \pm\sqrt{20} = \pm \sqrt{20}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =5 - \sqrt{20} = 0.528 s = 5 + \sqrt{20} = 9.472
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.