The number of real roots of (x+3)^4 + (x+5)^4 = 16 is:
a) 0
b) 2
c) 4
d) none of these

Question

The number of real roots of (x+3)^4 + (x+5)^4 = 16 is:
a) 0 
b) 2
c) 4
d) none of these

anonymous · Answer

2

king · Answer

hw?

anonymous · Answer

homework

king · Answer

how?

anonymous · Answer

-3 and -5
I did inspection that's all I did, but you should expand it if you can

king · Answer

what do u mean by u did inspection?

anonymous · Answer

you can't do it, not now, maybe when you will have enough experience then... but for now just expand it

anonymous · Answer

(x + 3)^4 = 4^2 - ((x +5)^2)^2

now try it

king · Answer

ok

anonymous · Answer

That's rite, even I did it by seeing it in first glimpse. It is a DPP question of Vidyamandir Classes, NewDelhi. Is der any formula for finding the direct roots of a quartic equation?

aravindg · Answer

hmm surprising i think i saw this qn before wait,,

anonymous · Answer

Really? Then please answer something. Please.

aravindg · Answer

w8

anonymous · Answer

Okay bhai!!!!!!

aravindg · Answer

hey i got it aadarsh!!!

aravindg · Answer

http://openstudy.com/updates/4f716097e4b07738f5adcf42

anonymous · Answer

I had posted the same question in maths group also. Got the answers.

king · Answer

the formula fr findin roots of a quadratic equation is
$$\frac$$

king · Answer

$$\frac{-b \ \pm \sqrt{b ^{2}-4ac}}{2a}$$

anonymous · Answer

$$(x+3)^4 + (x + 5)^4 = 16 $$$$x + 3 = t \tag1$$From (1),
$$t^4 + (t+2)^4 = 16$$$$t^4 + (t+2)^4 = 2^4$$One can easily make out the equation is valid only when t is either 0 or -2.$$t = -2,0$$From (1), $x = -5,-3$.