# (ii)let x be a +ve integer suh that unit digit of x is prime and product of all digits of x is also prime.How many such integers are possible (iii)number of real roots in x^4+(2-x)^4=34

1
by rajusetu
it was a silly mistake by me
sorry
sorry sir
sir x is a 4digit number.

2014-12-25T16:57:22+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
X = positive integer.  it is not said that  x is prime.

units digit of x  is prime,  so units digit  is  one of  2, 3, 5, 7.

product of digits of x is prime.  Only way it can happen is  that all other digits of x are all equal to 1.  Then the number can be
2, 3, 5,  7,  12, 13, 15, 17,    112, 113, 115, 117,    11112, 1113,  1115,  1117, ..
111112,  111113, 111115, 111117, ....

So the number such integers is infinite.
================================
the number of roots of
x^4 + (2 - x)^4 = 34

expand the polynomial and simply to get

P(x) =  x^4 - 4 x^3 + 12 x^2 - 16 x - 9 = 0
P(-x)  = (-x)^4 - 4 (-x)^3 + 12 (-x)^2 - 16 (-x) -9 = 0

Descartes rule.  It gives an upper bound on the number of positive, and negative real roots.  It does not necessarily give  the exact number of real roots.

coefficients in the polynomial are P(x) : +1, -4, +12, -16, -9
Sign of the coefficients changes plus to minus and minus to plus,  three times as we move from left to right.  So MAXimum possible real roots are 3.  There may be less than that.

Coefficients of P(-x) =  +1, +4, +12 , +16, -9
there is only one change in the sign, from +16 to -9.  So there is a MAXimum of one negative real root.

Maximum possible real roots are  3+1 = 4.  But it is possible that there are less number of real roots.  Minimum number of imaginary roots are 0.  It can be possibly more than 0, ie., 2 also.

If there are rational roots, they are possibly  1, -1, +3, -3, +9, or -9.
Trying them , P(x) is not 0.  There are no rational roots to this polynomial.

==================================
By examining the polynomial, and trying we get that
P(x) = (x^2 - 2x -1) (x^2  - 2x - 9)

the first quadratic expression has real roots : 1 + √2,  1 - √2

the second quadratic expression has imaginary roots: 1 + 2√2 i, 1 - 2√2 i

Actual number of real roots is 2.
=====================================
hands of clock
hour hand moves  360 degrees in  12 hours.  It moves 18 degrees in 18/360 * 12  = 3/5 hours = 36 minutes.

Minutes hand moves  360 degrees in  1 hour.  So it moves 6 degrees in one minute.  SO it moves  36 * 6 = 216 degrees.

click on thank you and select best answer
but answer was 4 roots were possibler