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x^{2}-5x+4=0
Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)^{2}.
a+b=-5 ab=4
To solve the equation, factor x^{2}-5x+4 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,-4 -2,-2
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 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-4 b=-1
The solution is the pair that gives sum -5.
\left(x-4\right)\left(x-1\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=4 x=1
To find equation solutions, solve x-4=0 and x-1=0.
x^{2}-5x+4=0
Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)^{2}.
a+b=-5 ab=1\times 4=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
-1,-4 -2,-2
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 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-4 b=-1
The solution is the pair that gives sum -5.
\left(x^{2}-4x\right)+\left(-x+4\right)
Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).
x\left(x-4\right)-\left(x-4\right)
Factor out x in the first and -1 in the second group.
\left(x-4\right)\left(x-1\right)
Factor out common term x-4 by using distributive property.
x=4 x=1
To find equation solutions, solve x-4=0 and x-1=0.
x^{2}-5x+4=0
Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)^{2}.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-16}}{2}
Multiply -4 times 4.
x=\frac{-\left(-5\right)±\sqrt{9}}{2}
x=\frac{-\left(-5\right)±3}{2}
Take the square root of 9.
x=\frac{5±3}{2}
The opposite of -5 is 5.
x=\frac{8}{2}
Now solve the equation x=\frac{5±3}{2} when ± is plus. Add 5 to 3.
x=4
Divide 8 by 2.
x=\frac{2}{2}
Now solve the equation x=\frac{5±3}{2} when ± is minus. Subtract 3 from 5.
x=1
Divide 2 by 2.
x=4 x=1
The equation is now solved.
x^{2}-5x+4=0
Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)^{2}.
x^{2}-5x=-4
Subtract 4 from both sides. Anything subtracted from zero gives its negation.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=\frac{9}{4}