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\sqrt{80x+89}=10+3x
Subtract -3x from both sides of the equation.
\left(\sqrt{80x+89}\right)^{2}=\left(10+3x\right)^{2}
Square both sides of the equation.
80x+89=\left(10+3x\right)^{2}
Calculate \sqrt{80x+89} to the power of 2 and get 80x+89.
80x+89=100+60x+9x^{2}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+3x\right)^{2}.
80x+89-100=60x+9x^{2}
Subtract 100 from both sides.
80x-11=60x+9x^{2}
Subtract 100 from 89 to get -11.
80x-11-60x=9x^{2}
Subtract 60x from both sides.
20x-11=9x^{2}
Combine 80x and -60x to get 20x.
20x-11-9x^{2}=0
Subtract 9x^{2} from both sides.
-9x^{2}+20x-11=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=20 ab=-9\left(-11\right)=99
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-11. To find a and b, set up a system to be solved.
1,99 3,33 9,11
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 99.
1+99=100 3+33=36 9+11=20
Calculate the sum for each pair.
a=11 b=9
The solution is the pair that gives sum 20.
\left(-9x^{2}+11x\right)+\left(9x-11\right)
Rewrite -9x^{2}+20x-11 as \left(-9x^{2}+11x\right)+\left(9x-11\right).
-x\left(9x-11\right)+9x-11
Factor out -x in -9x^{2}+11x.
\left(9x-11\right)\left(-x+1\right)
Factor out common term 9x-11 by using distributive property.
x=\frac{11}{9} x=1
To find equation solutions, solve 9x-11=0 and -x+1=0.
\sqrt{80\times \frac{11}{9}+89}-3\times \frac{11}{9}=10
Substitute \frac{11}{9} for x in the equation \sqrt{80x+89}-3x=10.
10=10
Simplify. The value x=\frac{11}{9} satisfies the equation.
\sqrt{80\times 1+89}-3=10
Substitute 1 for x in the equation \sqrt{80x+89}-3x=10.
10=10
Simplify. The value x=1 satisfies the equation.
x=\frac{11}{9} x=1
List all solutions of \sqrt{80x+89}=3x+10.