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Algebra

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26 Dec 2008 18:21:57 IST

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734

Quadratic equation....

None

If the polynomial P(x) = is a perfect square then (a+b) = ?

A-0

B-1

C-10

D-18

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Comments (9)

sarang

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26 Dec 2008 20:58:30 IST

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putting x=1
because product is one so definately 1 is its real root
0=1+1+a+b-8
0=a+b-6
a+b=6

Hari Shankar

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26 Dec 2008 21:52:44 IST

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a+b is not 6. Please try again

Apoorv

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26 Dec 2008 22:58:41 IST

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m gettin.....20.......

n hv verified it.........

Soumik

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26 Dec 2008 23:05:23 IST

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Is it multiple options correct?

Single option correct gives 0...

Anant Kumar

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27 Dec 2008 00:41:18 IST

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a+b=10.

Hari Shankar

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27 Dec 2008 05:37:58 IST

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or 22

Anant Kumar

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27 Dec 2008 06:38:34 IST

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Yes.

$P(x)=x^4 + ax^3 + bx^2 -8x +1$

Since $P(x)$ is a perfect square, we can write

$P(x)=(Ax^2+B x + C)^2=A^2x^4 + 2AB x^3 + (B^2 + 2AC)x^2+2BC x + C^2$

Comparing coefficients of like powers, we see that

$A^2=1$ , $2AB=a$ , $B^2+2BC = b$ , $2BC=-8$ , and $C^2=1$

From these relations, it is easy to see that $a+b =10$ or $a+b=22$

Hari Shankar

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27 Dec 2008 09:18:15 IST

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We can reduce the labour involved by noting that the leading coefficient i.e. coefficient of x⁴, and the constant term are both 1

So, either $x^4+ax^3+bx^2-8x+1 \equiv (x^2+cx+1)^2$ or

$x^4+ax^3+bx^2-8x+1 \equiv (x^2+cx-1)^2$

In the first case c = -4 so that a = -8 and b = 18 giving a+b = 10

In the second case c = 4, a = 8 and b = 14 giving a+b = 22

Rajat Khanduja

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27 Dec 2008 10:31:29 IST

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Can, we proceed this way.

If it is a perfect square, essentially, its roots have been reduced to 2 from 4.

Then, if p and q are the roots, then,

p.p.q.q = 1 , 2p+2q = -a , 2p.p + 2q.q + 4pq = b

and another equation , summantion of roots taken three at a time = 8

So we have 4 equations and 4 variables

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