If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less

Question

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

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nick1816 · Accepted Answer

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

ArunSharma12 · Answer

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

sathriyan · Answer

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

ArunSharma12 · Answer

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}

preetamsaha · Answer

nick1816  ii) f(3)>0 why is it?why not <0 ? what is the logic behind it ?

nick1816 · Answer

coefficient of x^2 is positive, so the equation is an upward parabola.

preetamsaha · Answer

nick1816  ok thanks. got it.

damd · Answer

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

Bunuel

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

Zuki1803 · Answer

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

nick1816 · Answer

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.

kelbyandrews · Answer

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!

Crytiocanalyst · Answer

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

devashish2407 · Answer

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?

Bunuel · Answer

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.

sravojitgoswami · Answer

f(x) = x2 - 2ax + a2 + a - 3 = 0; roots < 3 For roots to be real, b2 - 4ac > or = 0 Therefore, (-2a)2 - 4(1)(a2 + a - 3) >= 0 => 4a2 - 4a2 - 4a + 12 >= 0 => 4a =< 12 => a =< 3 On option evaluation, we see that option (c), (d), and (e) will be eliminated as the options contradict “a =< 3” On evaluating option (b), we cannot definitely conclude if a lies between 2 and 3. Hence, eliminated. Option (a) sits apt, as “a < 2” is definitely concurrent with “a =< 3”. Hence (a) is the answer. Please let me know if this particular approach is valid for this question?

AJSIPA · Answer

nick1816  can you elaborate why 2a<3+3 and not 2a<3? and why f(3)>0?

Kinshook · Answer

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? Sum of the roots = 2a < 6; a<3 f(3) > 0; 9 -6a + a^2 + a - 3 > 0; a^2-5a+6 > 0; (a-2)(a-3)>0; a>3 or a<2 Combining we get; a<2 IMO A

If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less

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If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards