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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.
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
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(-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
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}-5x+4-4=-4
Subtract 4 from both sides of the equation.
x^{2}-5x=-4
Subtracting 4 from itself leaves 0.
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}=-2.5. Then add the square of -\frac{5}{2}=-2.5 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}=-2.5 by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=\frac{9}{4}
\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}
Factor x^{2}-5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}
Simplify.
x=4 x=1
Add \frac{5}{2}=2.5 to both sides of the equation.
x ^ 2 -5x +4 = 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 = 5 rs = 4
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{5}{2} - u s = \frac{5}{2} + u
Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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{5}{2} - u) (\frac{5}{2} + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
\frac{25}{4} - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-\frac{25}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{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{5}{2} - \frac{3}{2} = 1 s = \frac{5}{2} + \frac{3}{2} = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.