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m^{2}-10m+13=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.
m=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 13}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
m=\frac{-\left(-10\right)±\sqrt{100-4\times 13}}{2}
Square -10.
m=\frac{-\left(-10\right)±\sqrt{100-52}}{2}
Multiply -4 times 13.
m=\frac{-\left(-10\right)±\sqrt{48}}{2}
Add 100 to -52.
m=\frac{-\left(-10\right)±4\sqrt{3}}{2}
Take the square root of 48.
m=\frac{10±4\sqrt{3}}{2}
The opposite of -10 is 10.
m=\frac{4\sqrt{3}+10}{2}
Now solve the equation m=\frac{10±4\sqrt{3}}{2} when ± is plus. Add 10 to 4\sqrt{3}.
m=2\sqrt{3}+5
Divide 10+4\sqrt{3} by 2.
m=\frac{10-4\sqrt{3}}{2}
Now solve the equation m=\frac{10±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 10.
m=5-2\sqrt{3}
Divide 10-4\sqrt{3} by 2.
m=2\sqrt{3}+5 m=5-2\sqrt{3}
The equation is now solved.
m^{2}-10m+13=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.
m^{2}-10m+13-13=-13
Subtract 13 from both sides of the equation.
m^{2}-10m=-13
Subtracting 13 from itself leaves 0.
m^{2}-10m+\left(-5\right)^{2}=-13+\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.
m^{2}-10m+25=-13+25
Square -5.
m^{2}-10m+25=12
Add -13 to 25.
\left(m-5\right)^{2}=12
Factor m^{2}-10m+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(m-5\right)^{2}}=\sqrt{12}
Take the square root of both sides of the equation.
m-5=2\sqrt{3} m-5=-2\sqrt{3}
Simplify.
m=2\sqrt{3}+5 m=5-2\sqrt{3}
Add 5 to both sides of the equation.
x ^ 2 -10x +13 = 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 = 13
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) = 13
To solve for unknown quantity u, substitute these in the product equation rs = 13
25 - u^2 = 13
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 13-25 = -12
Simplify the expression by subtracting 25 on both sides
u^2 = 12 u = \pm\sqrt{12} = \pm \sqrt{12}
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{12} = 1.536 s = 5 + \sqrt{12} = 8.464
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.