# (i)if x+1/x=66⇒50x/4x^2+36x+4 is what

2## Answers

### This Is a Certified Answer

=> 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

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.