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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
2 years, 4 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),

$$=\boxed{-126}$$ · 2 years, 4 months ago

Manipulation at its best. · 2 years, 3 months ago

$$x^{2} + 2x + 1 + 5 =0$$

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

$$x = -1 \pm i\sqrt{5}$$

( since any root is asked)

thus

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

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

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

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

$$= -126$$ · 2 years, 4 months ago

Just factorise it and substitute the value of 'r' in the quadratic equation obtained. · 2 years, 4 months ago

SORRY ITS-126 JUST SOLVED · 2 years, 4 months ago

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