মূল বিষয়বস্তুতে এড়িয়ে যান
$\exponential{x}{2} - 10 x + 25 = 0$
Solve for x
গ্রাফ

## শেয়ার করুন

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}
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.