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\left(\sqrt{2y+17}\right)^{2}=\left(y+9\right)^{2}
Square both sides of the equation.
2y+17=\left(y+9\right)^{2}
Calculate \sqrt{2y+17} to the power of 2 and get 2y+17.
2y+17=y^{2}+18y+81
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+9\right)^{2}.
2y+17-y^{2}=18y+81
Subtract y^{2} from both sides.
2y+17-y^{2}-18y=81
Subtract 18y from both sides.
-16y+17-y^{2}=81
Combine 2y and -18y to get -16y.
-16y+17-y^{2}-81=0
Subtract 81 from both sides.
-16y-64-y^{2}=0
Subtract 81 from 17 to get -64.
-y^{2}-16y-64=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-16 ab=-\left(-64\right)=64
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-64. To find a and b, set up a system to be solved.
-1,-64 -2,-32 -4,-16 -8,-8
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 64.
-1-64=-65 -2-32=-34 -4-16=-20 -8-8=-16
Calculate the sum for each pair.
a=-8 b=-8
The solution is the pair that gives sum -16.
\left(-y^{2}-8y\right)+\left(-8y-64\right)
Rewrite -y^{2}-16y-64 as \left(-y^{2}-8y\right)+\left(-8y-64\right).
y\left(-y-8\right)+8\left(-y-8\right)
Factor out y in the first and 8 in the second group.
\left(-y-8\right)\left(y+8\right)
Factor out common term -y-8 by using distributive property.
y=-8 y=-8
To find equation solutions, solve -y-8=0 and y+8=0.
\sqrt{2\left(-8\right)+17}=-8+9
Substitute -8 for y in the equation \sqrt{2y+17}=y+9.
1=1
Simplify. The value y=-8 satisfies the equation.
\sqrt{2\left(-8\right)+17}=-8+9
Substitute -8 for y in the equation \sqrt{2y+17}=y+9.
1=1
Simplify. The value y=-8 satisfies the equation.
y=-8 y=-8
List all solutions of \sqrt{2y+17}=y+9.