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a+b=-8 ab=15
To solve the equation, factor x^{2}-8x+15 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
-1,-15 -3,-5
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.
-1-15=-16 -3-5=-8
Calculate the sum for each pair.
a=-5 b=-3
The solution is the pair that gives sum -8.
\left(x-5\right)\left(x-3\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=5 x=3
To find equation solutions, solve x-5=0 and x-3=0.
a+b=-8 ab=1\times 15=15
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.
-1,-15 -3,-5
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.
-1-15=-16 -3-5=-8
Calculate the sum for each pair.
a=-5 b=-3
The solution is the pair that gives sum -8.
\left(x^{2}-5x\right)+\left(-3x+15\right)
Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).
x\left(x-5\right)-3\left(x-5\right)
Factor out x in the first and -3 in the second group.
\left(x-5\right)\left(x-3\right)
Factor out common term x-5 by using distributive property.
x=5 x=3
To find equation solutions, solve x-5=0 and x-3=0.
x^{2}-8x+15=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{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64-60}}{2}
Multiply -4 times 15.
x=\frac{-\left(-8\right)±\sqrt{4}}{2}
Add 64 to -60.
x=\frac{-\left(-8\right)±2}{2}
Take the square root of 4.
x=\frac{8±2}{2}
The opposite of -8 is 8.
x=\frac{10}{2}
Now solve the equation x=\frac{8±2}{2} when ± is plus. Add 8 to 2.
x=5
Divide 10 by 2.
x=\frac{6}{2}
Now solve the equation x=\frac{8±2}{2} when ± is minus. Subtract 2 from 8.
x=3
Divide 6 by 2.
x=5 x=3
The equation is now solved.
x^{2}-8x+15=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}-8x+15-15=-15
Subtract 15 from both sides of the equation.
x^{2}-8x=-15
Subtracting 15 from itself leaves 0.
x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-8x+16=-15+16
Square -4.
x^{2}-8x+16=1
Add -15 to 16.
\left(x-4\right)^{2}=1
Factor x^{2}-8x+16. 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-4\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x-4=1 x-4=-1
Simplify.
x=5 x=3
Add 4 to both sides of the equation.
x ^ 2 -8x +15 = 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 = 8 rs = 15
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 = 4 - u s = 4 + u
Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. 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>
(4 - u) (4 + u) = 15
To solve for unknown quantity u, substitute these in the product equation rs = 15
16 - u^2 = 15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 15-16 = -1
Simplify the expression by subtracting 16 on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =4 - 1 = 3 s = 4 + 1 = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.