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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
ArunSharma12 wrote:
If all the roots of the equation \(x2−2ax+a2+a−3=0\) are less than 3, then which of the following must be true?

A. a < 2
B. 2 < a < 3
C. 3 < a < 4
D. 4 < a < 5
E. a > 5

sum of the roots = 2a < 6; a < 3. ans seems to be between A & B.

on solving the equation, x = \(a ± \sqrt{3-a}\)

taking first root:
\(a-\sqrt{3-a}<3\)
\(\sqrt{3-a}(-1 - \sqrt{3-a}) < 0\)
\(\sqrt{3-a}(1+\sqrt{3-a}) > 0\)
solving inequality,
\(\sqrt{3-a} < -1\) not possible or
\(\sqrt{3-a} > 0; a < 3\)


taking second root:
\(a + \sqrt{3-a} < 3\)
\(\sqrt{3-a} < 3 -a\); as both sides are positive I can square both sides
\(3-a < (3-a)^2\)
\((3-a) - (3-a)^2 < 0\)
\((3-a)(a-2) < 0\); I know (3-a) is positive; (a-2) must be < 0 or a < 2

Ans: A


Hello !! Can you please elaborate how you have solved the equation to arrive the highlighted part?

Thanks in advance.. :)
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
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sathriyan wrote:
ArunSharma12 wrote:
If all the roots of the equation \(x2−2ax+a2+a−3=0\) are less than 3, then which of the following must be true?

on solving the equation, x = \(a ± \sqrt{3-a}\)

Hello !! Can you please elaborate how you have solved the equation to arrive the highlighted part?

Thanks in advance.. :)


applying the quadratic formula: \(x=\frac{-b±\sqrt{b^2-4a*c}}{2a}\)

\(x=\frac{-(-2a)±\sqrt{(-2a)^2-4(a^2+a-3)}}{2}\)
\(x=\frac{2a±\sqrt{4a^2-4a^2-4a+12}}{2}\)
\(x=a±\sqrt{-a+3}\)
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
nick1816 ii) f(3)>0 why is it?why not <0 ? what is the logic behind it ?
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
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coefficient of x^2 is positive, so the equation is an upward parabola.



preetamsaha wrote:
nick1816 ii) f(3)>0 why is it?why not <0 ? what is the logic behind it ?
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
nick1816 ok thanks. got it.
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
[quote="nick1816"]Since both roots are less than 3 and coefficient of x is positive,

i)Sum of roots < 3+3
2a<6
a<3

ii) f(3)>0

9-6a+a^2+a-3 >0
a^2-5a+6 >0
(a-2)(a-3)>0

a<2 or a>3 ( rejected, as 'a' can't be greater than 3.)


A


Hey, Can you explain to me on how you arrived at the below conclusion-

a<2 or a>3 ( rejected, as 'a' can't be greater than 3.)

Thanks
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
I took the 1,2 as maximum assumed roots if you add 1+2=3
hence, sum of roots = -b/a = 2a/1=2a
=2a<4
=a<2
product of roots = c/a =a^2+a-3
=a^2+a-3<3
=(a+3)(a-2)<0
=-3<a<2
union of both
a<2
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
damd
Product of 2 numbers is positive, if both of them are positive or negative.

(a-2)(a-3)>0

i) if a<2, both a-2 and a-3 are negative.
ii) if a>3, both a-2 and a-3 are positive.
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
roots of the equation :

2a +- sqrt ( 4a^2 - 4 ( a^2 + a - 3 ) )
______________________________ = 0 ( less than 3 )
2

so , a = 3/4 , -1
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
Can you just number pick and plug-in for a?

I started with 2 (since it would help me decide between A & B, and got X^2-4x+3 = (X-3)(X-1) and realized we can't have any root equal to 3. So a has to be less than 2.

Does this method work? If not, please explain!
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
kelbyandrews wrote:
Can you just number pick and plug-in for a?

I started with 2 (since it would help me decide between A & B, and got X^2-4x+3 = (X-3)(X-1) and realized we can't have any root equal to 3. So a has to be less than 2.

Does this method work? If not, please explain!


Even though it might feel like a brute force method i had a similar line of reasoning however used the magical number

let us assume a=0
=> x=+\sqrt{3} or -\sqrt{3}
which are both less than 3

therefore i safely concluded a<2 eleminating all other possibilites

Hence IMO A
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
Bunuel wrote:
If all the roots of the equation \(x^2 - 2ax + a^2 + a - 3 = 0\) are less than 3, then which of the following must be true?

A. a < 2
B. 2 < a < 3
C. 3 < a < 4
D. 4 < a < 5
E. a > 5


Are You Up For the Challenge: 700 Level Questions

­Bunuel

Hello, I have a doubt here. 

So in the question it is given that roots are less than 3. What does that mean? Does it mean that for some values of x the equation will be zero? Or Does it mean that for some values of a the equation will be zero? 

 
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
Expert Reply
devashish2407 wrote:
Bunuel wrote:
If all the roots of the equation \(x^2 - 2ax + a^2 + a - 3 = 0\) are less than 3, then which of the following must be true?

A. a < 2
B. 2 < a < 3
C. 3 < a < 4
D. 4 < a < 5
E. a > 5


Are You Up For the Challenge: 700 Level Questions

­Bunuel

Hello, I have a doubt here. 

So in the question it is given that roots are less than 3. What does that mean? Does it mean that for some values of x the equation will be zero? Or Does it mean that for some values of a the equation will be zero? 


­It would be clearer if the question specified that x is a variable and a is a constant. Regardless, "all the roots of the equation are less than 3" means that both values of x that satisfy the equation (the roots) are less than 3.
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Re: If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less [#permalink]
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