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Solution 2.3:2b

From Förberedande kurs i matematik 1

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m (Lösning 2.3:2b moved to Solution 2.3:2b: Robot: moved page)
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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:


y2+2y−15=y+12−12−15=y+12−16 

The equation can now be written as


y+12=16 


and has, after taking the square root, the solutions


y+1=16=4  which gives y=−1+4=3


y+1=−16=−4  which gives y=−1−4=−5


A quick check shows that y=−5 and y=3 satisfy the equation:


y=−5 : LHS= −52+2−5−15=25−10−15=0  = RHS

y=3 : LHS= 32+23−15=9+6−15=0 = RHS

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