Quadratic roots

Let \('r'\) be a root of the equation :

\(x^{2}+2x+6=0\)

Then find the value of: \((r+2)(r+3)(r+4)(r+5)\)

I wan't to know all possible approaches to this problem.

Would be gratified if got some responses :D


\(\text{DETAILS}\):

This appeared in KVPY SA this year(today :P)

Note by Aritra Jana
3 years, 7 months ago

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As \(r\) is the root of the given equation, \(r^{2}+2r+6=0\) ...... (1)

And

\((r+2)(r+3)(r+4)(r+5)\) = \((r+2)(r+5)(r+4)(r+3)\)

\(=({r}^{2}+7r+10)\) \((r^{2}+7r+12)\)

Substituting value of \(r^{2}\) from (1),

\(=(5r+4)(5r+6)\)

\(=(25{r}^{2}+50r+24)\)

Again substituting value of \(r^{2}\) from (1),

\(=\boxed{-126}\)

Karthik Sharma - 3 years, 7 months ago

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Manipulation at its best.

Krishna Ar - 3 years, 6 months ago

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\( x^{2} + 2x + 1 + 5 =0\)

\( (x + 1)^{2} = -5\)

\( x = -1 \pm i\sqrt{5}\)

( since any root is asked)

thus

\( (- 1 + i\sqrt{5} + 2)( - 1 + i\sqrt{5} + 3)( - 1 + i\sqrt{5} + 4)( - 1 + i\sqrt{5} + 5)\)

\(= (1 + i\sqrt{5})(2 + i\sqrt{5})(3 + i\sqrt{5})(4 + i\sqrt{5})\)

\( = (1 + i\sqrt{5})(4 + i\sqrt{5})(2 + i\sqrt{5})(3 + i\sqrt{5})\)

\( = ( -1 + 5i\sqrt{5})(1 + 5i\sqrt{5})\)

\( = -126\)

U Z - 3 years, 7 months ago

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Just factorise it and substitute the value of 'r' in the quadratic equation obtained.

Manmeswar Patnaik - 3 years, 7 months ago

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SORRY ITS-126 JUST SOLVED

Navdeep Nainwal - 3 years, 7 months ago

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