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22 Apr 2022, 00:33
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Which of the following is the product of two integers whose sum is 11?
(A) -42
(B) -28
(C) 12
(D) 26
(E) 32
As an alternate approach, why don't we use the concept of a quadratic equation to solve this question?
We know that an equation of the quadratic equation can be written as \(x^2\) - (sum of the roots )x + product of the roots = 0.
Here in this question, we will consider the two integers as the roots of the quadratic equation.
So it's given that the sum of the roots is 11 and the product of the roots is what we need to figure out.
\(x^2\) - 11x + product of the roots = 0
On comparing with the general form of the quadratic equation, \(ax^2 + bx + c =0\), we will get the values of a,b and c.
a =1, b= -11 , c= product of the roots.
We knew that the property of discriminant can be used to find the nature of the roots of a quadratic equation without actually solving the equation
The discriminant can be positive, negative, or zero.
If the discriminant is positive, the equation has two real and distinct roots. If the discriminant is zero, the equation has two equal and real roots. If the discriminant is negative, the equation has no real solution.
Since the roots of this quadratic equation here is an integer, we can conclude that the Discriminant is positive and should be a perfect square. If it's not a perfect square, then the roots will not be an integer, but irrational roots. We can use this as the initial criteria to eliminate the answer options. If this initial condition is satisfied, as a final check, we can see if \((-b ± \sqrt{ Discriminant} )/2a\) is an integer or not. This step is only required if 2 or more answer options satisfy the initial condition that the Discriminant should be a perfect square.
Discriminant = \(b^2 - 4ac\) should be a perfect square.
==> \((-11)^2 - 4*1*c \)= 121 - 4c should be a perfect square, where c is the product of the roots.
Next Step: plugin answer choices instead of c and check if 121 - 4c is a perfect square or not.
A. -42 ===> 121 - 4*-42 = 121 + 168 = 289 . Bingo
!! Its a perfect square. \(17^2 = 289\) Lucky that you got it in the first attempt.
This is the first condition for getting the roots as an integer. Next step, either you can check if any other options will also lead to a perfect square and eliminate them accordingly or confirm that the roots here are integers by checking the second condition.
\((-b ± \sqrt{ Discriminant}) /2a\) should be an integer.
\((11 ± \sqrt{289} )/2\) i.e two roots are (11 + 17)/2 = 14 and (11-17)/2 = -3
Since both the roots are integers, we can confirm that option A is the answer.
In case of option A is not giving a perfect square, you should follow the same process for other answer options
(B) -28===> 121 - 4*-28= 121 + 112 = 233 . This is not a perfect square. Eliminated
(C) 12 ===> 121 - 4*12= 121 - 48 = 73 This is not a perfect square. Eliminated
(D) 26===> 121 - 4*26 = 121 - 104 = 17 This is not a perfect square. Eliminated
(E) 32===> 121 - 4*32 = 121 - 129 = -8 The Discriminant is negative, hence eliminated.
I hope the explanation is clear.
Thanks,
Clifin J Francis,
GMAT Mentor.
(A) -42
(B) -28
(C) 12
(D) 26
(E) 32
As an alternate approach, why don't we use the concept of a quadratic equation to solve this question?
We know that an equation of the quadratic equation can be written as \(x^2\) - (sum of the roots )x + product of the roots = 0.
Here in this question, we will consider the two integers as the roots of the quadratic equation.
So it's given that the sum of the roots is 11 and the product of the roots is what we need to figure out.
\(x^2\) - 11x + product of the roots = 0
On comparing with the general form of the quadratic equation, \(ax^2 + bx + c =0\), we will get the values of a,b and c.
a =1, b= -11 , c= product of the roots.
We knew that the property of discriminant can be used to find the nature of the roots of a quadratic equation without actually solving the equation
The discriminant can be positive, negative, or zero.
If the discriminant is positive, the equation has two real and distinct roots. If the discriminant is zero, the equation has two equal and real roots. If the discriminant is negative, the equation has no real solution.
Since the roots of this quadratic equation here is an integer, we can conclude that the Discriminant is positive and should be a perfect square. If it's not a perfect square, then the roots will not be an integer, but irrational roots. We can use this as the initial criteria to eliminate the answer options. If this initial condition is satisfied, as a final check, we can see if \((-b ± \sqrt{ Discriminant} )/2a\) is an integer or not. This step is only required if 2 or more answer options satisfy the initial condition that the Discriminant should be a perfect square.
Discriminant = \(b^2 - 4ac\) should be a perfect square.
==> \((-11)^2 - 4*1*c \)= 121 - 4c should be a perfect square, where c is the product of the roots.
Next Step: plugin answer choices instead of c and check if 121 - 4c is a perfect square or not.
A. -42 ===> 121 - 4*-42 = 121 + 168 = 289 . Bingo
!! Its a perfect square. \(17^2 = 289\) Lucky that you got it in the first attempt.This is the first condition for getting the roots as an integer. Next step, either you can check if any other options will also lead to a perfect square and eliminate them accordingly or confirm that the roots here are integers by checking the second condition.
\((-b ± \sqrt{ Discriminant}) /2a\) should be an integer.
\((11 ± \sqrt{289} )/2\) i.e two roots are (11 + 17)/2 = 14 and (11-17)/2 = -3
Since both the roots are integers, we can confirm that option A is the answer.
In case of option A is not giving a perfect square, you should follow the same process for other answer options
(B) -28===> 121 - 4*-28= 121 + 112 = 233 . This is not a perfect square. Eliminated
(C) 12 ===> 121 - 4*12= 121 - 48 = 73 This is not a perfect square. Eliminated
(D) 26===> 121 - 4*26 = 121 - 104 = 17 This is not a perfect square. Eliminated
(E) 32===> 121 - 4*32 = 121 - 129 = -8 The Discriminant is negative, hence eliminated.
I hope the explanation is clear.
Thanks,
Clifin J Francis,
GMAT Mentor.
15 Aug 2022, 06:24
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took me 50 seconds. Here is my approach:
a + b = 11 -> ab = ?
quick mental math
1 + 10 -> 10 ?
2 + 9 -> 18 ?
3 + 8 -> 24?
4 + 7 -> 28?
5 + 6 -> 30?
6 + 5 -> repetiton
7 + 4
....
Next, move to one negative integer (it does not matter which, the product will be same)
-1 + 12 -> -12?
-2 + 13 -> -26?
-3 + 14 -> -42 YES (A)
a + b = 11 -> ab = ?
quick mental math
1 + 10 -> 10 ?
2 + 9 -> 18 ?
3 + 8 -> 24?
4 + 7 -> 28?
5 + 6 -> 30?
6 + 5 -> repetiton
7 + 4
....
Next, move to one negative integer (it does not matter which, the product will be same)
-1 + 12 -> -12?
-2 + 13 -> -26?
-3 + 14 -> -42 YES (A)
22 Aug 2022, 22:46
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a+b=11
a*b=42
a*(11-a) = 42
a^2 - 11a + 42 = 0
(a-14)(a+3) --> a=14 is a valid solution. therefore, b=-3 .... a+b = 14-3 = 11 and a*b=(14)*(-3) = -42
a*b=42
a*(11-a) = 42
a^2 - 11a + 42 = 0
(a-14)(a+3) --> a=14 is a valid solution. therefore, b=-3 .... a+b = 14-3 = 11 and a*b=(14)*(-3) = -42
