# (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

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by rajusetu
See at question ,Is this (x+1)/66 and 50x/(4x^2+36x+4).?
or, x+1/x and 50x/(4x^2+36x+4)?
x+1/x and 50x/(4x^2+36x+4)
please write one query in one question. it will be easier to answer
create a separate question for the (ii) part . bye

## Answers

The Brainliest Answer!
2014-12-23T17:31:11+05:30

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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

=====================

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

reply
i am sorry. i will not answer. u wrote so many parts in the question.
sir i gave as separate question
then will you naswer
plz reply
• Brainly User
2014-12-24T22:26:14+05:30

1. You have, and can be written as .
From this you get, .

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,
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, , 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 .
Roots of this equation are -10 & -3.

I'm pretty sure about my answers.
And the answer for the second question is not 4 as you can see that there are more than 6 numbers like that. See the answer. All of them have primes in their unit's place & their product is also prime.
You may use a graph generator for the third one and verify.
For the third, two roots are real and the other two are complex.
And one thing complex roots always exist in conjugate pairs for polynomials of even degree.