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If (a + b) = 4, and ab = 7/4, what is the value of a^2 – b^2?
A. -12
B. 12
C. Either A or B
D. 16
E. None
A. -12
B. 12
C. Either A or B
D. 16
E. None
GMATWhizTeam
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Expert reply
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It is given that (a + b) = 4 and ab = \(\frac{7}{4}\).
The value of \(a^2\) – \( b^2\) is to be calculated. Using standard algebraic identities, \(a^2\) – \(b^2\) = (a + b) (a – b)
Since the value of (a + b) is already given, we need the value of (a – b) to calculate the value of the given expression.
Using standard algebraic identities, \((a – b) ^2\) = \((a + b)^2\) – 4ab.
Substituting the values, \((a -b)^2\) = \((4)^2\) – 4 * \(\frac{7}{4}\) = 16 – 7 = 9
Therefore, (a – b) = 3 or (a – b) = -3
If (a – b) = 3, the value of \(a^2\) – \(b^2\) = 4 * 3 = 12.
If (a – b) = -3, the value of \(a^2\) – \(b^2\) = 4 * -3 = -12.
Therefore, the value of \(a^2\) – \(b^2\) could be 12 or -12.
The correct answer option is C.
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The value of \(a^2\) – \( b^2\) is to be calculated. Using standard algebraic identities, \(a^2\) – \(b^2\) = (a + b) (a – b)
Since the value of (a + b) is already given, we need the value of (a – b) to calculate the value of the given expression.
Using standard algebraic identities, \((a – b) ^2\) = \((a + b)^2\) – 4ab.
Substituting the values, \((a -b)^2\) = \((4)^2\) – 4 * \(\frac{7}{4}\) = 16 – 7 = 9
Therefore, (a – b) = 3 or (a – b) = -3
If (a – b) = 3, the value of \(a^2\) – \(b^2\) = 4 * 3 = 12.
If (a – b) = -3, the value of \(a^2\) – \(b^2\) = 4 * -3 = -12.
Therefore, the value of \(a^2\) – \(b^2\) could be 12 or -12.
The correct answer option is C.
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General Discussion
I'm getting answer D...
ab = 7/4
(a+b)^2 = 4^2
a^2 + 2ab + b^2 = 16
(a+b) = 4
7/4b + b = 4
28b + 4b^2 = 16b
4b^2 + 12b = 0
b^2 + 3b = 0
(b+3)*(b-0) = 0
So b is either -3 or 0...
Plugging in these values to a+b = 4...
If b is 0, then:
a+0 = 4
a = 4
If b is 3, then:
a+3 = 4
a = 1
So a^2 - b^2...
It's either:
1^2 - 3^2 = 1-9 = -8
Or...
4^2 - 0^2 = 16
16 (answer D) has to be the answer.
ab = 7/4
(a+b)^2 = 4^2
a^2 + 2ab + b^2 = 16
(a+b) = 4
7/4b + b = 4
28b + 4b^2 = 16b
4b^2 + 12b = 0
b^2 + 3b = 0
(b+3)*(b-0) = 0
So b is either -3 or 0...
Plugging in these values to a+b = 4...
If b is 0, then:
a+0 = 4
a = 4
If b is 3, then:
a+3 = 4
a = 1
So a^2 - b^2...
It's either:
1^2 - 3^2 = 1-9 = -8
Or...
4^2 - 0^2 = 16
16 (answer D) has to be the answer.
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fmcgee1
I think you've got a little boo boo in the lines in red. Great example of why I frequently (usually?) try to avoid the "real" math...all it takes is one silly mistake and you're in trouble!
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Asked: If (a + b) = 4, and ab = 7/4, what is the value of a^2 – b^2?
a = 7/4b
7/4b + b = 4
7 + 4b^2 - 16b = 0
4b^2 - 14b - 2b + 7 = 0
(2b-1)(2b-7) = 0
b = 1/2 or 7/2
a = 7/2 or 1/2
a^2 - b^2 = 1/4 - 49/4 = - 48/4 = -12
or
a^2 - b^2 = 49/4 - 1/4 = 48/4 = 12
IMO C
a = 7/4b
7/4b + b = 4
7 + 4b^2 - 16b = 0
4b^2 - 14b - 2b + 7 = 0
(2b-1)(2b-7) = 0
b = 1/2 or 7/2
a = 7/2 or 1/2
a^2 - b^2 = 1/4 - 49/4 = - 48/4 = -12
or
a^2 - b^2 = 49/4 - 1/4 = 48/4 = 12
IMO C
GMATWhizTeam
Hi I have a doubt
if we sqaure up a+b=4 then we get
a2 +2(7/4) +b2 =16
a2 +b2 =16-7/2
a2+b2=25/2
cant we substitute a2=12.5-b2 or b2=12.5-a2?
anushridi
That would lead you to a^2 - b^2 = a^2 - (12.5 - a^2) = 2a^2 - 12.5. However, we need to find numerical values of a^2 – b^2, not the value of it in terms of a or b.
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Hello Bunuel and KarishmaB
Could you please tell me where I am going wrong:
Given that
( a + b ) = 4..........(i) and
ab = 7/4,
We need to find a^2 - b^2, i.e, (a+b)(a-b).
Now, ab = 7/4,
so 2ab = 14/4.......(ii)
(I) - (ii)
a+b - 2ab = 4 - 14/4
(a - b)^2 = 2/4 = 1/2
Thus (a-b) = (+/- [square_root 1/2])....(iii)
Now multiplying (i) by (iii)
(a + b) (a - b) = 4 * (+/- [square_root 1/2]).
Thank you for your time and effort!
Could you please tell me where I am going wrong:
Given that
( a + b ) = 4..........(i) and
ab = 7/4,
We need to find a^2 - b^2, i.e, (a+b)(a-b).
Now, ab = 7/4,
so 2ab = 14/4.......(ii)
(I) - (ii)
a+b - 2ab = 4 - 14/4
(a - b)^2 = 2/4 = 1/2
Thus (a-b) = (+/- [square_root 1/2])....(iii)
Now multiplying (i) by (iii)
(a + b) (a - b) = 4 * (+/- [square_root 1/2]).
Thank you for your time and effort!
rishabhm99
The red part is not correct.
a + b - 2ab is not the same as (a - b)^2, which equals a^2 + b^2 - 2ab.
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I have solved it using an alternate approach. The detailed solution is provided in the video.
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Shivanshsm
Posts: 3
I think the answer should be B,
because a+b=4 i.e. positive and a.b=7/4 i.e. positive so from a*b= positive we can conclude either both are positive or both are negative. If we take the case where a & b both are negative so a+b will lso be negative which is not the case, hence a & b should both be positive.
Please let me know if I am doing something wrong.
because a+b=4 i.e. positive and a.b=7/4 i.e. positive so from a*b= positive we can conclude either both are positive or both are negative. If we take the case where a & b both are negative so a+b will lso be negative which is not the case, hence a & b should both be positive.
Please let me know if I am doing something wrong.
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A different way...
\(a^2 - b^2 = (a-b) * (a+b) = 4(a-b)\)
\((a+b) = 4\)
\((a*b) = \frac{7}{4}\)
\(b = \frac{7}{(4a)}\)
\(\frac{7}{(4a)}+a = 4\)
\(a = \frac{(4+\sqrt{3})}{2}\)
or
\(a = \frac{(4-\sqrt{3})}{2}\)
Plug into \(a^2 - b^2 = (a-b) * (a+b) = 4(a-b)\)
To get -12 or 12.
I think I could have just stopped when realized that two options would be possible, however answer E leaves a lot of space for a wrong assumption...
\(a^2 - b^2 = (a-b) * (a+b) = 4(a-b)\)
\((a+b) = 4\)
\((a*b) = \frac{7}{4}\)
\(b = \frac{7}{(4a)}\)
\(\frac{7}{(4a)}+a = 4\)
\(a = \frac{(4+\sqrt{3})}{2}\)
or
\(a = \frac{(4-\sqrt{3})}{2}\)
Plug into \(a^2 - b^2 = (a-b) * (a+b) = 4(a-b)\)
To get -12 or 12.
I think I could have just stopped when realized that two options would be possible, however answer E leaves a lot of space for a wrong assumption...
It's a "lucky" question for me if I encounter it in the exam. Here is how I process
2 number, with mutiple is fraction, and sum is an integer, i immediately think about x.5 + y.5
and 7/4 = 1/2 * 7/2
1/2 + 7/2 = 4
))
So
a^2 - b^2 = (7/2)^2 - (1/2)^2 = 48/4 =12
and reverse we have -12
2 number, with mutiple is fraction, and sum is an integer, i immediately think about x.5 + y.5
and 7/4 = 1/2 * 7/2
1/2 + 7/2 = 4
))So
a^2 - b^2 = (7/2)^2 - (1/2)^2 = 48/4 =12
and reverse we have -12
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