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From Kaplan Diag. test equation

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isisalaska Community Manager Default Avatar
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From Kaplan Diag. test equation

Post Wed Mar 21, 2007 6:45 am
For the equation x^2 + 2x + m = 5, where m is a constant, 3 is one solution for x. What is the other solution?

a) -5
b) -2
c) -1
d0 3
E) 5

What do you do? do you subtiutute 3 where x is? I thought it was 5...

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Isis Alaska

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Post Fri Sep 29, 2017 2:30 pm
since 3 is one of the solutions of the quadratic equation, we will represent the whole solution with a letter say 'y'.
x=3 or x=y
and x-3=0 or x-y=0
(x-3)(x-y)=0
expanding the equation above, we have
x(x-y) - 3(x-y) = 0
x^2 - xy-3x+ 3y =0
x^2 - (y+3)x + 3y =0
Comparing this newly derived form of the equation with the original question, we observe that:
-(y+3)=2 ------------(1)
-y-3=2
-y=2+3=5
-y=5
y=-5
Hence, we obtained the other solution of the equation as required. option A is correct

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Cybermusings Legendary Member Default Avatar
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Post Tue Mar 27, 2007 9:55 am
For the equation x^2 + 2x + m = 5, where m is a constant, 3 is one solution for x. What is the other solution?

a) -5
b) -2
c) -1
d0 3
E) 5

x^2 + 2x + m = 5

3 is one solution for x

Thus the quadratic equation has to look like

x^2 + 2x + m - 5 = 0

For 3 to be one root it has to look like this

x^2 - 3x + x + m - 5 = 0

Therefore even m - 5 = -3

Therefore m = -3 + 5 = 2

Thus x^2 - 3x + x + 2 - 5

= x ^2 - 3x + x - 3

= x(x-3) + 1(x-3)

Thus the other root of the equation is -1

I aint too good at explaining but hope this helps

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Post Fri Sep 29, 2017 2:30 pm
since 3 is one of the solutions of the quadratic equation, we will represent the whole solution with a letter say 'y'.
x=3 or x=y
and x-3=0 or x-y=0
(x-3)(x-y)=0
expanding the equation above, we have
x(x-y) - 3(x-y) = 0
x^2 - xy-3x+ 3y =0
x^2 - (y+3)x + 3y =0
Comparing this newly derived form of the equation with the original question, we observe that:
-(y+3)=2 ------------(1)
-y-3=2
-y=2+3=5
-y=5
y=-5
Hence, we obtained the other solution of the equation as required. option A is correct

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