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a+b=-15 ab=44
To solve the equation, factor x^{2}-15x+44 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,-44 -2,-22 -4,-11
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 44.
-1-44=-45 -2-22=-24 -4-11=-15
Calculate the sum for each pair.
a=-11 b=-4
The solution is the pair that gives sum -15.
\left(x-11\right)\left(x-4\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=11 x=4
To find equation solutions, solve x-11=0 and x-4=0.
a+b=-15 ab=1\times 44=44
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+44. To find a and b, set up a system to be solved.
-1,-44 -2,-22 -4,-11
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 44.
-1-44=-45 -2-22=-24 -4-11=-15
Calculate the sum for each pair.
a=-11 b=-4
The solution is the pair that gives sum -15.
\left(x^{2}-11x\right)+\left(-4x+44\right)
Rewrite x^{2}-15x+44 as \left(x^{2}-11x\right)+\left(-4x+44\right).
x\left(x-11\right)-4\left(x-11\right)
Factor out x in the first and -4 in the second group.
\left(x-11\right)\left(x-4\right)
Factor out common term x-11 by using distributive property.
x=11 x=4
To find equation solutions, solve x-11=0 and x-4=0.
x^{2}-15x+44=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(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 44}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and 44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-15\right)±\sqrt{225-4\times 44}}{2}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-176}}{2}
Multiply -4 times 44.
x=\frac{-\left(-15\right)±\sqrt{49}}{2}
Add 225 to -176.
x=\frac{-\left(-15\right)±7}{2}
Take the square root of 49.
x=\frac{15±7}{2}
The opposite of -15 is 15.
x=\frac{22}{2}
Now solve the equation x=\frac{15±7}{2} when ± is plus. Add 15 to 7.
x=11
Divide 22 by 2.
x=\frac{8}{2}
Now solve the equation x=\frac{15±7}{2} when ± is minus. Subtract 7 from 15.
x=4
Divide 8 by 2.
x=11 x=4
The equation is now solved.
x^{2}-15x+44=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}-15x+44-44=-44
Subtract 44 from both sides of the equation.
x^{2}-15x=-44
Subtracting 44 from itself leaves 0.
x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=-44+\left(-\frac{15}{2}\right)^{2}
Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-15x+\frac{225}{4}=-44+\frac{225}{4}
Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-15x+\frac{225}{4}=\frac{49}{4}
Add -44 to \frac{225}{4}.
\left(x-\frac{15}{2}\right)^{2}=\frac{49}{4}
Factor x^{2}-15x+\frac{225}{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-\frac{15}{2}\right)^{2}}=\sqrt{\frac{49}{4}}
Take the square root of both sides of the equation.
x-\frac{15}{2}=\frac{7}{2} x-\frac{15}{2}=-\frac{7}{2}
Simplify.
x=11 x=4
Add \frac{15}{2} to both sides of the equation.
x ^ 2 -15x +44 = 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 = 15 rs = 44
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 = \frac{15}{2} - u s = \frac{15}{2} + u
Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{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>
(\frac{15}{2} - u) (\frac{15}{2} + u) = 44
To solve for unknown quantity u, substitute these in the product equation rs = 44
\frac{225}{4} - u^2 = 44
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 44-\frac{225}{4} = -\frac{49}{4}
Simplify the expression by subtracting \frac{225}{4} on both sides
u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{15}{2} - \frac{7}{2} = 4 s = \frac{15}{2} + \frac{7}{2} = 11
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.