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Steps Using Factoring
Steps Using Factoring By Grouping
Steps Using the Quadratic Formula
Steps for Completing the Square
Steps Using Direct Factoring Method
Steps Using Factoring
To solve the equation, factor using formula . To find and , set up a system to be solved.
Since is positive, and have the same sign. Since is negative, and are both negative. List all such integer pairs that give product .
Calculate the sum for each pair.
The solution is the pair that gives sum .
Rewrite factored expression using the obtained values.
Rewrite as a binomial square.
To find equation solution, solve .
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a+b=-10 ab=25
To solve the equation, factor x^{2}-10x+25 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,-25 -5,-5
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 25.
-1-25=-26 -5-5=-10
Calculate the sum for each pair.
a=-5 b=-5
The solution is the pair that gives sum -10.
\left(x-5\right)\left(x-5\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
\left(x-5\right)^{2}
Rewrite as a binomial square.
x=5
To find equation solution, solve x-5=0.
a+b=-10 ab=1\times 25=25
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+25. To find a and b, set up a system to be solved.
-1,-25 -5,-5
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 25.
-1-25=-26 -5-5=-10
Calculate the sum for each pair.
a=-5 b=-5
The solution is the pair that gives sum -10.
\left(x^{2}-5x\right)+\left(-5x+25\right)
Rewrite x^{2}-10x+25 as \left(x^{2}-5x\right)+\left(-5x+25\right).
x\left(x-5\right)-5\left(x-5\right)
Factor out x in the first and -5 in the second group.
\left(x-5\right)\left(x-5\right)
Factor out common term x-5 by using distributive property.
\left(x-5\right)^{2}
Rewrite as a binomial square.
x=5
To find equation solution, solve x-5=0.
x^{2}-10x+25=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(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-100}}{2}
Multiply -4 times 25.
x=\frac{-\left(-10\right)±\sqrt{0}}{2}
Add 100 to -100.
x=-\frac{-10}{2}
Take the square root of 0.
x=\frac{10}{2}
The opposite of -10 is 10.
x=5
Divide 10 by 2.
x^{2}-10x+25=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.
\left(x-5\right)^{2}=0
Factor x^{2}-10x+25. 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-5\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-5=0 x-5=0
Simplify.
x=5 x=5
Add 5 to both sides of the equation.
x=5
The equation is now solved. Solutions are the same.
x ^ 2 -10x +25 = 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 = 10 rs = 25
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 = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>
(5 - u) (5 + u) = 25
To solve for unknown quantity u, substitute these in the product equation rs = 25
25 - u^2 = 25
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 25-25 = 0
Simplify the expression by subtracting 25 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = 5
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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