Question

The number of real roots (x+3)^4+(x+5)^4=16

Answer 1

Given : (x+3)^4+(x+5)^4=16

We can write (x+3) as (x + 4 -1 )

Also we can write (x+5) as (x + 4 + 1)

We put (x + 4) = t

then, (x+3) becomes (t - 1)

and (x+5) becomes (t + 1)

So, we can rewrite the above equation as

(t-1)^4 + (t+1)^4 = 16

or, (t^4-4t^3+6t^2-4t+1) + (t^4+4t^3+6t^2+4t+1)
= 16

{expansion using binomial expansion }

or, 2t^4 + 12t^2 +2 = 16

or, 2(t^4 + 6t^2 + 1) = 16

or, t^4 + 6t^2 + 1 = 16/2 = 8

or, t^4 + 6t^2 -7 = 0

or, t^4 - t^2 + 7t^2 -7 = 0

or, t^2(t^2 - 1) +7(t^2 - 1) = 0

or, (t^2+7)(t^2-1) = 0

so, either (t^2+7) =0 or (t^2 -1 )= 0

For, t^2 +7 = 0

so, t^2 = -7

or, t = +√7i and -√7i __________is imaginary roots

For, t^2 - 1 = 0

so, t^2 = 1

or, t = +1 and -1 ___________real roots.

As, t = x + 4

so, 1 = x + 4

or, x = 1-4 = -3

Also, -1 = x + 4

so, x = -1-4= -5

So, the no. of real roots are two (2) as -5 and -3.

Answer 2

As we know that $(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$

Given: $(x+3)^4+(x+5)^4=16$

$(x+3)^4+(x+5)^5=16\\x^4+4x^3.3+6x^23^2+4x3^3+3^4+x^4+4x^35+6x^25^2+4x5^3+5^4=16\\x^4+12x^3+54x^2+108x+81+x^4+20x^3+150x^2+500x+625=16\\2x^4+32x^3+204x^2+608x+706=16\\2x^4+32x^3+204x^2+608x+890=0\\x^4+16x^3+102x^2+304x+445=0$

The equation has 4 real roots.

#SPJ2