Solve for x
x=5
x=48
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a+b=-53 ab=240
To solve the equation, factor x^{2}-53x+240 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,-240 -2,-120 -3,-80 -4,-60 -5,-48 -6,-40 -8,-30 -10,-24 -12,-20 -15,-16
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 240.
-1-240=-241 -2-120=-122 -3-80=-83 -4-60=-64 -5-48=-53 -6-40=-46 -8-30=-38 -10-24=-34 -12-20=-32 -15-16=-31
Calculate the sum for each pair.
a=-48 b=-5
The solution is the pair that gives sum -53.
\left(x-48\right)\left(x-5\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=48 x=5
To find equation solutions, solve x-48=0 and x-5=0.
a+b=-53 ab=1\times 240=240
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+240. To find a and b, set up a system to be solved.
-1,-240 -2,-120 -3,-80 -4,-60 -5,-48 -6,-40 -8,-30 -10,-24 -12,-20 -15,-16
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 240.
-1-240=-241 -2-120=-122 -3-80=-83 -4-60=-64 -5-48=-53 -6-40=-46 -8-30=-38 -10-24=-34 -12-20=-32 -15-16=-31
Calculate the sum for each pair.
a=-48 b=-5
The solution is the pair that gives sum -53.
\left(x^{2}-48x\right)+\left(-5x+240\right)
Rewrite x^{2}-53x+240 as \left(x^{2}-48x\right)+\left(-5x+240\right).
x\left(x-48\right)-5\left(x-48\right)
Factor out x in the first and -5 in the second group.
\left(x-48\right)\left(x-5\right)
Factor out common term x-48 by using distributive property.
x=48 x=5
To find equation solutions, solve x-48=0 and x-5=0.
x^{2}-53x+240=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(-53\right)±\sqrt{\left(-53\right)^{2}-4\times 240}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -53 for b, and 240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-53\right)±\sqrt{2809-4\times 240}}{2}
Square -53.
x=\frac{-\left(-53\right)±\sqrt{2809-960}}{2}
Multiply -4 times 240.
x=\frac{-\left(-53\right)±\sqrt{1849}}{2}
Add 2809 to -960.
x=\frac{-\left(-53\right)±43}{2}
Take the square root of 1849.
x=\frac{53±43}{2}
The opposite of -53 is 53.
x=\frac{96}{2}
Now solve the equation x=\frac{53±43}{2} when ± is plus. Add 53 to 43.
x=48
Divide 96 by 2.
x=\frac{10}{2}
Now solve the equation x=\frac{53±43}{2} when ± is minus. Subtract 43 from 53.
x=5
Divide 10 by 2.
x=48 x=5
The equation is now solved.
x^{2}-53x+240=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}-53x+240-240=-240
Subtract 240 from both sides of the equation.
x^{2}-53x=-240
Subtracting 240 from itself leaves 0.
x^{2}-53x+\left(-\frac{53}{2}\right)^{2}=-240+\left(-\frac{53}{2}\right)^{2}
Divide -53, the coefficient of the x term, by 2 to get -\frac{53}{2}. Then add the square of -\frac{53}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-53x+\frac{2809}{4}=-240+\frac{2809}{4}
Square -\frac{53}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-53x+\frac{2809}{4}=\frac{1849}{4}
Add -240 to \frac{2809}{4}.
\left(x-\frac{53}{2}\right)^{2}=\frac{1849}{4}
Factor x^{2}-53x+\frac{2809}{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{53}{2}\right)^{2}}=\sqrt{\frac{1849}{4}}
Take the square root of both sides of the equation.
x-\frac{53}{2}=\frac{43}{2} x-\frac{53}{2}=-\frac{43}{2}
Simplify.
x=48 x=5
Add \frac{53}{2} to both sides of the equation.
x ^ 2 -53x +240 = 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 = 53 rs = 240
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{53}{2} - u s = \frac{53}{2} + u
Two numbers r and s sum up to 53 exactly when the average of the two numbers is \frac{1}{2}*53 = \frac{53}{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{53}{2} - u) (\frac{53}{2} + u) = 240
To solve for unknown quantity u, substitute these in the product equation rs = 240
\frac{2809}{4} - u^2 = 240
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 240-\frac{2809}{4} = -\frac{1849}{4}
Simplify the expression by subtracting \frac{2809}{4} on both sides
u^2 = \frac{1849}{4} u = \pm\sqrt{\frac{1849}{4}} = \pm \frac{43}{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{53}{2} - \frac{43}{2} = 5 s = \frac{53}{2} + \frac{43}{2} = 48
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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