Factor X 2 X 20
$\exponential{x}{two} + x = 20 $
10=-5
x=four
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x^{ii}+x-20=0
Decrease 20 from both sides.
a+b=1 ab=-twenty
To solve the equation, factor ten^{two}+x-twenty using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, prepare a system to be solved.
-1,20 -2,x -4,5
Since ab is negative, a and b take the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.
-1+20=nineteen -2+10=eight -4+v=1
Calculate the sum for each pair.
a=-four b=5
The solution is the pair that gives sum i.
\left(x-4\right)\left(x+5\right)
Rewrite factored expression \left(x+a\right)\left(ten+b\right) using the obtained values.
x=4 x=-5
To find equation solutions, solve x-4=0 and 10+5=0.
10^{2}+x-twenty=0
Subtract 20 from both sides.
a+b=1 ab=1\left(-xx\right)=-twenty
To solve the equation, cistron the left manus side by grouping. First, left manus side needs to be rewritten as x^{two}+ax+bx-twenty. To find a and b, prepare up a system to be solved.
-i,20 -2,10 -four,5
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -twenty.
-1+xx=19 -two+10=8 -iv+5=i
Summate the sum for each pair.
a=-4 b=v
The solution is the pair that gives sum 1.
\left(x^{two}-4x\right)+\left(5x-20\right)
Rewrite x^{two}+x-20 as \left(10^{2}-4x\right)+\left(5x-20\right).
10\left(x-four\right)+5\left(ten-4\right)
Cistron out 10 in the first and 5 in the second grouping.
\left(10-4\right)\left(10+5\right)
Gene out mutual term x-iv by using distributive property.
x=4 x=-5
To detect equation solutions, solve x-four=0 and x+5=0.
x^{2}+x=20
All equations of the class ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives ii solutions, one when ± is improver and 1 when it is subtraction.
x^{2}+x-20=twenty-20
Subtract twenty from both sides of the equation.
x^{2}+ten-20=0
Subtracting 20 from itself leaves 0.
10=\frac{-1±\sqrt{1^{2}-4\left(-20\right)}}{2}
This equation is in standard form: ax^{two}+bx+c=0. Substitute 1 for a, 1 for b, and -twenty for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{ane-iv\left(-twenty\correct)}}{2}
Square one.
x=\frac{-ane±\sqrt{ane+80}}{ii}
Multiply -4 times -xx.
x=\frac{-ane±\sqrt{81}}{2}
Add 1 to lxxx.
ten=\frac{-1±9}{2}
Accept the foursquare root of 81.
x=\frac{viii}{2}
At present solve the equation 10=\frac{-1±9}{ii} when ± is plus. Add -i to nine.
10=\frac{-ten}{ii}
Now solve the equation x=\frac{-1±9}{two} when ± is minus. Subtract 9 from -i.
ten=four x=-5
The equation is at present solved.
x^{2}+ten=twenty
Quadratic equations such as this one can be solved past completing the square. In lodge to complete the square, the equation must first be in the class 10^{2}+bx=c.
x^{ii}+x+\left(\frac{i}{2}\right)^{2}=20+\left(\frac{1}{2}\right)^{two}
Divide i, the coefficient of the x term, by ii to get \frac{1}{2}. And then add the square of \frac{ane}{2} to both sides of the equation. This pace makes the left hand side of the equation a perfect square.
x^{2}+ten+\frac{1}{4}=20+\frac{ane}{four}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{ii}+x+\frac{1}{four}=\frac{81}{4}
Add 20 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{ii}=\frac{81}{4}
Factor ten^{2}+x+\frac{1}{4}. In general, when x^{two}+bx+c is a perfect square, it can always be factored equally \left(x+\frac{b}{two}\right)^{2}.
\sqrt{\left(10+\frac{one}{2}\right)^{2}}=\sqrt{\frac{81}{iv}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{9}{2} ten+\frac{i}{2}=-\frac{nine}{2}
Simplify.
x=4 ten=-five
Subtract \frac{one}{two} from both sides of the equation.
Factor X 2 X 20,
Source: https://mathsolver.microsoft.com/en/solve-problem/x%20%5E%20%7B%202%20%7D%20+%20x%20=%2020
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