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805+ (Hard)|   Algebra|      
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siddharthkapoor
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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = Updated on: 28 Oct 2019, 22:58
Originally posted by siddharthkapoor on 12 Oct 2019, 06:28.
Last edited by chetan2u on 28 Oct 2019, 22:58, edited 1 time in total.
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95% (hard)

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33% (02:51) correct 67% (02:30) wrong based on 90 sessions

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If a^2 and b^2 are integers and \((a^4 – a^2 + 4b^2 – 6)^{\frac{1}{2}} = 2\), what is the value of \(a^2\) and \(b^2\)?

(1) \(a^2\) = \((15 -2a^2)^{\frac{1}{2}}\)

(2) \((b^4 – 1)^{\frac{1}{2}}\) = 0
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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = 12 Oct 2019, 10:45
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iamsiddharthkapoor
If a^2 and b^2 are integers and \((a^4 – a^2 + 4b^2 – 6)^{1/2} = 2\), what is the value of \(a^2\) and \(b^2\)?

(1) \(a^2 = (15 -2a^2)^{1/2}\)

(2) \((b^4 – 1)^{1/2}\) = 0
Show more

from statement (1), \(a^4 + 2a^2 - 15 = 0\) , or \((a^2+5)(a^2-3) = 0\)
because because of the even power of \(a^2\), \(a^2\) can't be negative, so \(a^2 = 3\)
then \(a^2 = 3\) can be substituted into the original equation to find \(b^2\) (=1) --> sufficient

from statement (2), \(b^2 = 1\), and by substituting it in the original equation, \(a^4 -a^2 - 6 = 0\) , or \((a^2+2)(a^2-3) = 0\)
and again, because because of the even power of \(a^2\), \(a^2\) can't be negative, so \(a^2 = 3\) --> sufficient
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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = 12 Oct 2019, 12:58
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iamsiddharthkapoor
If a^2 and b^2 are integers and \((a^4 – a^2 + 4b^2 – 6)^(1/2) = 2\), what is the value of \(a^2\) and \(b^2\)?

(1) \(a^2\) = \((15 -2a^2)^(1/2)\)

(2) \((b^4 – 1)^(1/2)\) = 0
Show more

To find the values of two unknowns, we need two equations or the value of one unknown. Here, from the question, we can get one equation, \(a^4-a^2+4b^2 = 4\). We need one more equation.

Statement 1 gives us \(a^4 = 15- 2a^2\), another equation. SUFFICIENT.

Statement 2: by removing the square root, we get \(b^4 -1\) = 0, another equation. SUFFICIENT.

Answer D.
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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = 28 Oct 2019, 22:46
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Given , (a4–a2+4b2–6)(1/2)=2 and a^2, b^2 are integers.
so, a^4 - a^2 4*b^2 -6 = 4----> a^4-a^2+4*b^2 = 10
now comes with additional information -
1) a2 = (15−2a2)(1/2)----> a^4 = 15 - 2*a^2 -----> a^4+2*a^2-15 =0 -----> (a^2-3)(a^2+5)= 0
hence a^2 =3 and -5 , but square always positive. so a^2 = 3 only possible value.
now put a^2 =3 in a^4-a^2+4*b^2 = 10 ---> 9-3-10 +4*b^2 =0------> 4*b^2 = 4 ----->b^2= 1
hence this information is sufficient, so either A or D is correct., so C,B,E out. let's check second information.

2) (b4–1)(1/2)= 0------>b^4 =1 ---->b^2 =1, -1, but square always positive , so b^2 = 1.
now put b^2=1 in a^4-a^2+4*b^2 = 10 ----> a^4-a^2+4=10---->a^4-a^2-6 =0 ---->(a^2-3)(a^2+2)=0---->a^2=3,-2
we know square always positive. hence a^2 =3
this information is also sufficient.
Hence option D is correct.
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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = 28 Oct 2019, 23:07
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iamsiddharthkapoor
If a^2 and b^2 are integers and \((a^4 – a^2 + 4b^2 – 6)^{\frac{1}{2}} = 2\), what is the value of \(a^2\) and \(b^2\)?

(1) \(a^2\) = \((15 -2a^2)^{\frac{1}{2}}\)

(2) \((b^4 – 1)^{\frac{1}{2}}\) = 0
Show more

\((a^4 – a^2 + 4b^2 – 6)^{\frac{1}{2}} = 2\)
Square both sides and taking a^2=x and b^2=y... \(a^4 – a^2 + 4b^2 – 6 = 4.....x^2-x+4y-10=0\), where \(x\geq{0}\) and \(y\geq{0}\)



(1) \(a^2\) = \((15 -2a^2)^{\frac{1}{2}}\)
\(a^2\) = \((15 -2a^2)^{\frac{1}{2}}\).... \(x\) = \((15 -2x)^{\frac{1}{2}}\)
Square both sides...\(x^2=15-2x.......x^2+2x-15=0......(x+5)(x-3)=0\)
So x cannot be negative, so \(x=a^2=3\)
\(x^2-x+4y-10=0.........3^2-3+4y-10=0......4y=4...y=b^2=1\)
Suff

(2) \((b^4 – 1)^{\frac{1}{2}}\) = 0
Square both sides...
\(y^2-1=0.......(y-1)(y+1)=0\)
so y=-1 or 1. Again y cannot be negative, so \(y=b^2=1\) and \(x=a^2=3\)
Suff

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If a^2 and b^2 are integers and (a^4 – a^2 + 4b^2 – 6)^(1/2) = 03 Aug 2023, 07:57
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