Question

(i)if x+1/x=66⇒50x/4x^2+36x+4 is what
(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
(iv)the coefficient of x in eqn x^2+px+q=0 was taken 17 in place of 13 and its roots were found to be -2 and -15.The original roots are

Answer 1

     x + 1/ x = 66
      =>  x^2 + 1  =   66 x

Substitute this in the second equation,
    50 x / [ 4 (x^2 + 9 x + 1 ] =  50 x / [ 4 ( 66 x + 9 x ]      =  1 / 6

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  2 x^4 - 4 x^3 *2 + 6 x^2 * 2^2 - 4 x * 2^3 + 2^4  = 34
 P(x)  =  x^4 - 4 x^3 + 12 x^2 - 16 x - 18  = 0

 find the change in the signs of coefficients in P(x)  to know the number of positive and negative real roots.  This is given by Descartes' rule.

  coefficients of P(x) are +1, -4, +12, -16, -18 
   maximum num of positive real roots = 3  as sign changes  3 times.

P(-x) = x^4 + 4 x^3 + 12 x^2 + 16 x - 18
      there is only one sign change.  so maximum number of negative real roots is 1.

       total number of max real roots is 4.

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   coefficient of x is p.  its correct value is 13.   But initially p was taken wrongly as 17. 

  using    x^2 + 17 x + q = 0,  roots found are -2 and -15.
         q = product or roots = -2 * -15 = 30

  Hence the correct equation is obtained by substituting correct values of p and q  as :
         x^2 + 13 x + 30 = 0
             roots are  -3 and -10  by factorizing the quadratic expression.
                      (x + 3) (x + 10) = 0

Answer 2

1. You have, $x+ \frac{1}{x}=66$ and $\frac{50x}{4x^{2}+36x+4}$ can be written as $\frac{50}{4x+36+4\frac{1}{x}}$.
From this you get, $\frac{50}{4(66)+36}=\frac{1}{6}$ .

2. There are infinitely many numbers. Some of them are :- 2, 3, 5, 7, 13, 17, 113......

3. There are three different ways which I know to solve this problem.
→ 1. Find local points of local extrema, sort them up, substitute in the given expression and find how many times the expression changes its sign from -∞ to +∞.(generally used for polynomials of lower degree)
→ 2. Descartes's method. (Simplified form of the method mentioned above, usually used when the expression has simplified terms)
→ 3. We have something that says, " For a polynomial f(x) if f(a).f(b)<0 then there exists a real root between 'a' and 'b' ". ( I am gonna apply this. I admit it's time-consuming but you are 13 and might not probably understand the above two methods )
Here, $f(x)=x^{4}+(2-x)^{4}-34$
Now, start from -∞, you can see that expression is positive.
Take -1, even now f(x)>0.
Take 0, f(x)<0, observe that sign of f(x) has changed when moving from -1 to 0, this means they're is a root between -1 & 0.
Take 1, f(x)<0, no root between 0 & 1.
Take 3, f(x)>0, there's another root.
You may verify that expression remains positive from here on, indicating there no root between 3 and +∞.
Hence, there are only two real roots possible and the other two are complex.

4. You have, $x^{2}+px+q=0$, first p was noted to be 17 and the roots were found to be -2 & -15.
You get, q=(-2)(-15)=30. Then p was corrected as 13. New equation you get is $x^{2}+13x+30=0$.
Roots of this equation are -10 & -3.