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Difficulty:
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17% (02:56) correct
83%
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If \(a^2\) is an integer and \(\sqrt{a^6-a^4-b-1} =10\), what are the values of \(a^2\) and b?
(1) \(a^4=a^2+20\)
(2) Product of \(b\) and \(b-1\) is an even prime number.
(1) \(a^4=a^2+20\)
(2) Product of \(b\) and \(b-1\) is an even prime number.
30 Dec 2022, 13:19
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Let a^2=x
Statement I : a^4=a^2+20
Let a^2=x
x^2-x=20
x^2-x-20=0
x^2-5x+4x-20=0
(x-5)(x+4)=0
x=5 or x=-4
Since a square cannot be a negative number, reject a^2=-4
So, a^2= 5
Now put this value of a^2 in q.stem to derive value of b
125-25-b-1=100
b=-1
Statement I alone is sufficient.
Statement II: Product of b and (b-1) and is an even prime number.
b(b-1)=2
b can be 2 since 2(2-1)=2
OR b can be -1 since -1(-1-1)= (-1)(-2)=2
If b=2, a^6-a^4-2-1=100
a^6-a^4= 103
103 is a prime number
b cannot be 2.
b=-1
a^6-a^a4-(-1)-1=100
a^6-a^a4+1-1=100
a^6-a^a4=100
a^4(a^2-1)= 100
Let a^2 =x
x^2(x-1)=100
25(5-1)=100
x= 5
a^2=5
Statement II alone is also sufficient.
Option D is the answer
Posted from my mobile device
Statement I : a^4=a^2+20
Let a^2=x
x^2-x=20
x^2-x-20=0
x^2-5x+4x-20=0
(x-5)(x+4)=0
x=5 or x=-4
Since a square cannot be a negative number, reject a^2=-4
So, a^2= 5
Now put this value of a^2 in q.stem to derive value of b
125-25-b-1=100
b=-1
Statement I alone is sufficient.
Statement II: Product of b and (b-1) and is an even prime number.
b(b-1)=2
b can be 2 since 2(2-1)=2
OR b can be -1 since -1(-1-1)= (-1)(-2)=2
If b=2, a^6-a^4-2-1=100
a^6-a^4= 103
103 is a prime number
b cannot be 2.
b=-1
a^6-a^a4-(-1)-1=100
a^6-a^a4+1-1=100
a^6-a^a4=100
a^4(a^2-1)= 100
Let a^2 =x
x^2(x-1)=100
25(5-1)=100
x= 5
a^2=5
Statement II alone is also sufficient.
Option D is the answer
Posted from my mobile device
05 Mar 2024, 05:33
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Hi divyadna, in a^6-a^4= 103 , why can't that subtraction be a prime number?
