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17 Jul 2011 01:25:17 IST

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the number of real roots of the equation (x^2 + 2x)^2 - (x + 1)^2 - 55 =0 is a)2 b)1 c)4 d)none of t

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the number of real roots of the equation (x^2 + 2x)^2 - (x + 1)^2 - 55 =0 is a)2 b)1 c)4 d)none of these.....please show the calculation also...thank u.

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MAYANK KUMAR JHA

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17 Jul 2011 01:48:42 IST

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the given equation which can be rearranged as (4x^2-1)(x+1)^2-55=0....now it being a polynomial(continous) of four degree. let f(x)=(4x^2-1)(x+1)^2-55......which implies f'(x)=8x^3+12x^2+3x-1=0...........f''(x)=24x^2+24x+3...f'' has only

single root which implies by the converse of

rolle's theorem that f'(x) wud have atleast

tw roots........and further by converse of

rolle's theorem we have f(x) to have atleast

three real roots.........................

Now since complex roots always exist in conjugates provided the coefficients are real.......therefore thr fourth root must also be real....and hence we have 'c' as the correct answer.......

Aakash Verma

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17 Jul 2011 02:10:29 IST

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but the answer given is b)1 ......that's y i'm confused which one is correct.

MAYANK KUMAR JHA

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17 Jul 2011 03:45:59 IST

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dude the option 'b' is undoubtedly wrong.......consult the source once more or tell it..............as complex roots for a polynomial having real coefficients must exist in pairs.........

Hari Shankar

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17 Jul 2011 07:58:57 IST

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hemang

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17 Jul 2011 19:54:16 IST

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see is simple.

we take (x+1)² = t

then the qeuation becomes,

(t-1)² - t - 55 = 0

so we get t = 9 and -6

but t is a whole square so it can't be negative.

we reject t = -6

so t = 9

or, (x+1)² = 9

therefore (x+1) = +3 or -3

so x = -4 or x = 2.

two solutions exist.

Aakash Verma

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17 Jul 2011 22:10:26 IST

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thanks a lot. :)

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