Solution 2.3:2b

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The first step when we solve the second-degree equation is to complete the square on the left-hand side:
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<center> [[Image:2_3_2b.gif]] </center>
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<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>
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The equation can now be written as
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<math>\left( y+1 \right)^{2}=16</math>
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and has, after taking the square root, the solutions
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<math>y+1=\sqrt{16}=4</math>
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which gives
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<math>y=-1+4=3</math>
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<math>y+1=-\sqrt{16}=-4</math>
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which gives
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<math>y=-1-4=-5</math>
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A quick check shows that
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<math>y=-\text{5 }</math>
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and
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<math>y=\text{3 }</math>
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satisfy the equation:
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<math>y=-\text{5 }</math>: LHS=
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<math>\left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0</math>
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= RHS
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<math>y=\text{3 }</math>: LHS=
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<math>3^{2}+2\centerdot 3-15=9+6-15=0</math>
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= RHS

Revision as of 13:31, 20 September 2008

The first step when we solve the second-degree equation is to complete the square on the left-hand side:


\displaystyle y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.

The equation can now be written as


\displaystyle \left( y+1 \right)^{2}=16


and has, after taking the square root, the solutions


\displaystyle y+1=\sqrt{16}=4 which gives \displaystyle y=-1+4=3


\displaystyle y+1=-\sqrt{16}=-4 which gives \displaystyle y=-1-4=-5


A quick check shows that \displaystyle y=-\text{5 } and \displaystyle y=\text{3 } satisfy the equation:


\displaystyle y=-\text{5 }: LHS= \displaystyle \left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0 = RHS

\displaystyle y=\text{3 }: LHS= \displaystyle 3^{2}+2\centerdot 3-15=9+6-15=0 = RHS

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