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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.