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a+b=-54 ab=104
To solve the equation, factor x^{2}-54x+104 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,-104 -2,-52 -4,-26 -8,-13
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 104.
-1-104=-105 -2-52=-54 -4-26=-30 -8-13=-21
Calculate the sum for each pair.
a=-52 b=-2
The solution is the pair that gives sum -54.
\left(x-52\right)\left(x-2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=52 x=2
To find equation solutions, solve x-52=0 and x-2=0.
a+b=-54 ab=1\times 104=104
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+104. To find a and b, set up a system to be solved.
-1,-104 -2,-52 -4,-26 -8,-13
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 104.
-1-104=-105 -2-52=-54 -4-26=-30 -8-13=-21
Calculate the sum for each pair.
a=-52 b=-2
The solution is the pair that gives sum -54.
\left(x^{2}-52x\right)+\left(-2x+104\right)
Rewrite x^{2}-54x+104 as \left(x^{2}-52x\right)+\left(-2x+104\right).
x\left(x-52\right)-2\left(x-52\right)
Factor out x in the first and -2 in the second group.
\left(x-52\right)\left(x-2\right)
Factor out common term x-52 by using distributive property.
x=52 x=2
To find equation solutions, solve x-52=0 and x-2=0.
x^{2}-54x+104=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(-54\right)±\sqrt{\left(-54\right)^{2}-4\times 104}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -54 for b, and 104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-54\right)±\sqrt{2916-4\times 104}}{2}
Square -54.
x=\frac{-\left(-54\right)±\sqrt{2916-416}}{2}
Multiply -4 times 104.
x=\frac{-\left(-54\right)±\sqrt{2500}}{2}
Add 2916 to -416.
x=\frac{-\left(-54\right)±50}{2}
Take the square root of 2500.
x=\frac{54±50}{2}
The opposite of -54 is 54.
x=\frac{104}{2}
Now solve the equation x=\frac{54±50}{2} when ± is plus. Add 54 to 50.
x=52
Divide 104 by 2.
x=\frac{4}{2}
Now solve the equation x=\frac{54±50}{2} when ± is minus. Subtract 50 from 54.
x=2
Divide 4 by 2.
x=52 x=2
The equation is now solved.
x^{2}-54x+104=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}-54x+104-104=-104
Subtract 104 from both sides of the equation.
x^{2}-54x=-104
Subtracting 104 from itself leaves 0.
x^{2}-54x+\left(-27\right)^{2}=-104+\left(-27\right)^{2}
Divide -54, the coefficient of the x term, by 2 to get -27. Then add the square of -27 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-54x+729=-104+729
Square -27.
x^{2}-54x+729=625
Add -104 to 729.
\left(x-27\right)^{2}=625
Factor x^{2}-54x+729. 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-27\right)^{2}}=\sqrt{625}
Take the square root of both sides of the equation.
x-27=25 x-27=-25
Simplify.
x=52 x=2
Add 27 to both sides of the equation.
x ^ 2 -54x +104 = 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 = 54 rs = 104
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 = 27 - u s = 27 + u
Two numbers r and s sum up to 54 exactly when the average of the two numbers is \frac{1}{2}*54 = 27. 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>
(27 - u) (27 + u) = 104
To solve for unknown quantity u, substitute these in the product equation rs = 104
729 - u^2 = 104
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 104-729 = -625
Simplify the expression by subtracting 729 on both sides
u^2 = 625 u = \pm\sqrt{625} = \pm 25
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =27 - 25 = 2 s = 27 + 25 = 52
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.