If a1 and a2 are the real roots of x^2-px+12=0, then which of the

Question

If  a_1  and  a_2  are the real roots of  x^2-px+12=0 , then which of the following statements is definitely true?

A.  |a_1+a_2| \leq 2 \sqrt{3}

B.  |a_1-a_2| \leq 2 \sqrt{3}

C.  |a_1+a_2| \geq 4 \sqrt{3}

D.  |a_1-a_2| \geq 4 \sqrt{3}

E. None of the above

Source: CAT 2019 India

darshshah981 · Accepted Answer

x^2-px+12=0

Lets look at this quadratic equation is the form of  ax^2+bx+c=0 
We know that sum of the roots of a quadratic is  -b/a  and the product of the roots of a quadratic is  c/a

Now, if the roots are real then the discriminant  \geq  0, that is  b^2 - 4ac   \geq  0

>>  p^2 - 4(1)(12)   \geq  0
>>  p^2   \geq  48
>> | p |  \geq   \sqrt{48} 
>> | p |  \geq   4\sqrt{3}

But p is the sum of the roots (a1+a2)...(Since sum of the roots of a quadratic equation is  -b/a )

Therefore, | a1+a2 |  \geq  4 \sqrt{3}  or C is the correct answer.

globaldesi · Answer

sum of roots should be greater than 0
hence  p^2 >48
or  p>4\sqrt{3} 
now p = sum of roots
hence C is the correct answer

TestPrepUnlimited · Answer

In the GMAT, there is almost always a way to solve questions regarding quadratics without using the quadratic formula.

For all quadratics in this form  x^2 - ax + b = 0 ,  a  is the sum of roots and  b  is the product of roots. Therefore we know  a1*a2 = 12  and  a1 + a2 = p .
If both roots are positive, we can minimize  p  when  a1 = a2 = \sqrt{12} , so that  p = a1 + a2 = 2\sqrt{12} = 4\sqrt{3} , which is a minimum so we have  a1 + a2 >= 4\sqrt{3} . We can do the opposite with negative roots to get a similar result, regardless all signs point to C.

Crytiocanalyst · Answer

first let us simply apply a plugging in anf verification 
let the roots be 3 and 4 
it satisfies only the third equation

and another method being if real roots exists then b^2-4ac>=0
p^2-48>=0
=>|a1+a2|=4(3)^1/2 since sum of roots is p 
Therefore IMO C

BrushMyQuant · Answer

a_1  and  a_2  are the real roots of  x^2-px+12=0

=> Determinant  ≥ 0
=> If equation is ax^2 + bx + c = 0
Then Determinant =  b^2  - 4ac ≥ 0

=>  (-p)^2  - 4 * 1 * 12 ≥ 0
=>  p^2  - 48 ≥ 0
=>  p^2  ≥ 48
=> |p| ≥  \sqrt{48} 
=> |p| ≥ 4 \sqrt{3}

Sum of roots  =  a_1  +  a_2  =  \frac{-b}{a}  =  \frac{-(-p)}{1}  = p
=> |p| = |  a_1  +  a_2  | ≥ 4 \sqrt{3}

So,  Answer will be C
 Hope it helps!

Watch the following video to MASTER Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

arushi118 · Answer

SUm of roots of a quadratic = -b/a
product of roots of a quadratic = c/a

for real roots: D>=0 -> p^2 - 4(12)>=0 -> p>=4root(3)
So, sum of roots = |a1+a2| = p >=4root(3)
C is correct.

If a1 and a2 are the real roots of x^2-px+12=0, then which of the

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If a1 and a2 are the real roots of x^2-px+12=0, then which of the

Prep Toolkit

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