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If y= (x-1)(x+2), then what is the least possible value of y?

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If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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If y= (x-1)(x+2), then what is the least possible value of y?

a. -3
b. -9/4
c. -2
d. -3/2
e. 0
[Reveal] Spoiler: OA
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If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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New post 21 Jun 2017, 13:25
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If y= (x-1)(x+2) = x^2 + x - 2

Now when
x=0, y = 0 + 0 - 2 = -2
x=-1/2, y=1/4 - 1/2 - 2 = -9/4 = -2.25
x=-1/3, y=1/9 - 1/3 - 2 = -8/3 = -2.67
x=-1/4, y=1/16 -1/4 - 2 = -35/16 = -2.18
x=-1, y = 1 - 1 -2 = -2

Of the available options, Option B is smallest possible value given in the answer options
The least value of y occurs when x=-1/2
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If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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we get the equation for y as y = x^2 +x-2
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B
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Re: If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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New post 21 Jun 2017, 21:15
y = (x-1)(x+2) = x^2 + x - 2 = x^2 + x + (1/4 - 1/4) - 2 (adding 1/4 and subtracting 1/4 makes no difference)

= (x^2 + x + 1/4) - 1/4 - 2

Now x^2 + x + 1/4 is square of (x + 1/2). So we have

(x+1/2)^2 - 9/4

We can see that the square term (x+1/2)^2 can only take a minimum value of 0, not less than that. In that case the expression
0^2 - 9/4 = -9/4

Hence B answer
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Re: If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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New post 27 Aug 2017, 03:21
umadurga wrote:
we get the equation for y as y = x^2 +x-2
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B


Got to dust off my old math equations. The least value one equation -b/2a is the most useful in this question.

Thanks !!!!!!
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Re: If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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New post 27 Aug 2017, 19:46
umadurga wrote:
we get the equation for y as y = x^2 +x-2
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B

how do you get a=1, b=1, c=-2 tho?
from a equation like this --> y=x^2+ x^-2
please advise, thank you
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If y= (x-1)(x+2), then what is the least possible value of y? [#permalink]

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New post 27 Aug 2017, 22:37
Arsh4MBA wrote:
umadurga wrote:
we get the equation for y as y = x^2 +x-2
This is in the form ax^2 +Bx +c, the least value of the expression is when x=-b/2a
here a=1, b=1 and c=-2
Least value for y is when x=-b/2a => = -1/2
Substituting this value
y=(-1/2)^2+(-1/2)-2
=1/4-1/2-2
=-1/4-2
=-9/4
Option B


Got to dust off my old math equations. The least value one equation -b/2a is the most useful in this question.

Thanks !!!!!!

pclawong
(x-1)(x+2)= x^2+2x-x-2
=x^2+x-2

This is a quadratic equation of the form ax^2+bx+c
Hence a=1, b=1, c=-2

Minimum value of x= -b/2a =-1/2
If y= (x-1)(x+2), then what is the least possible value of y?   [#permalink] 27 Aug 2017, 22:37
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