# Quadratic Equations based on x + 1/x type and other algebraic identities

Thursday, August 20th, 2020 ## Quadratic Equations based on x + 1/x:

x + 1/x is itself a category of questions. There are many variants of this in which either you were given a value of this equation or you have to find the higher degree value which would be similar to this equation.
For a clear understanding let’s do some examples:

Example 1: If 6(x2 + 1/x2) – 5(x + 1/x) = 38 and x is positive and rational then find the value of x.

Sol: In this question, firstly we have to make the first bracket as a complete square of the second bracket. This we can by adding 2.x.1/x which is equivalent to 2. Then the equation becomes:

6(x2 + 1/x2 +2) –  5(x + 1/x) = 50 { 38 + 6*2)

⇒ 6(x2 + 1/x2 +2) –  5(x + 1/x) – 50 = 0

Now put x + 1/x = y

⇒  6y2 -5y -50 = 0

⇒  (2y +5)(3y-10)= 0

⇒  y=-5/2 or 10/3

As x is positive therefore, x + 1/x =10/3

On solving further you will get as x=3 or 1/3

Example 2: If x + 2/x = √6, then find the value of x6 + x12 + x18.

Sol: Given x + 2/x = √6

Cubing both sides,

x^3 + 8/x^3 + 3(x)(2/x)(x +2/x) = 6√6

x^3 + 8/x^3 + 3(2)(√6)= 6√6 ⇒ x^3+ 8/x^3 =0

Multiplying both sides by x^3, x6 =-8

x^6+x^12+x^18 =x^6+(x^6 )^2+(x^6 )^3
=-8 +(-8)^2 + (-8)^(3) =-456

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Example 3: If x + 1/x = -√3, then find the value of x42 + x48 + x54 + x60 + x66 + x72.

Sol:  Given x + 1/x = -√3

Cubing both sides,

x^3 + 1/x^3 + 3(x)(1/x)(x +1/x) = -3√3

x^3 + 1/x^3 + 3(-√3)= -3(-√3)

x^3 + 1/x^3 =0

Multiplying both sides by x3, x6 = -1

x^42+x^48+x^54+ x^60 + x^66 + x^72 = (x^6 )^7 +(x^6 )^8+ (x^6 )^9+ (x^6 )^10 + (x^6 )^11 +(x^6 )^12

= (-1)^7+(-1)^8+(-1)^9+(-1)^10+(-1)^11+(-1)^12

= 0

Example 4: If x + 1/x = 3 , then find the value of x6+1/x6.

Sol:  Given : (x + 1/x) = 3

On, squaring both the sides we get

(x + 1/x)2 = 32
x2 +(1/x2) + 2 = 9
⇒  x2 +(1/x2)  = 7

Now cubing on both the sides we get,
[x2 +(1/x2)]3  = 73

LHS is in the form of (a + b)3 = a3 + b3 + 3ab (a + b)

Hence [x2]3 + [(1/x2)]3 + 3 (x2) × (1/ x2)[ x2 +(1/ x2)]  = 343

⇒ x6 + (1/x6) + 3 × 7 = 343
⇒ x6 + (1/x6) + 21 = 343
∴  x6 + (1/x6)  = 343 − 21 = 322

Example 5: If x +1/x =5, what does 1/x2 + x2 equal?

Sol: Given : x +1/x =5

⇒  1/x = 5-x

So, 1 = 5*x – x2

⇒  x2 – 5x + 1 = 0

Clearly, the discriminant is not a perfect square, therefore we have to solve via quadratic formula

x = [5 ± √21]/2

so

either x = 5/2 + √21/2

or x = 5/2 – √21/2

Let’s work with the solution x = 5/2 + √21/2

Now we compute x2 + 1/x2

First of all, x2 = 25/4 + 5*√21/2 + 21/4

= 46/4 + 5*√21/2

= [23 + 5*√21)]/2

Thus

1/x2 + x2 = 2/[23 + 5*√21)] + [23 + 5*√21)]/2

Obtain a common denominator:

4/(2*[23 + 5*√21]) + [23 + 5*√21)]^2/(2*[23 + 5*√21])

=  4 + [23 + 5*√21]2

———————– =

2*[23 + 5*√21]

4 + 529 + 230*√21 + 525

—————————- =

46 + 10*√21

1058 + 230*√21

——————-

46 + 10*√21

23*(46 + 10*√21)

⇒  ——————–

(46 + 10*√21)

= 23

If we had chosen instead to work with the solution x = 5/2 – √21/2, the computations would have been different, but the answer comes out the same.  Perhaps you can try it on your own.

## Incorrect Coefficients Questions:

In this type of questions you were given two situations or two persons. In the first situation, in finding the roots, the person has written the constant term incorrectly and in the second situation, the person has written the coefficient of x incorrect. Then you have to find either the correct equation or corrects roots.

It can be clearly understood by this example:

Example 1: John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

Sol: John had made mistake while writing the constant term, therefore, he must have written the coefficient of x correct i.e. the sum of the roots must be correct.

Therefore sum of roots = 5+9 = 14

Jane made a mistake in writing the coefficient of x and got roots as 12 and 4, so she must have written the constant term correct i.e. the products of the roots would be correct.

Thus, products of roots= 12 x 4 = 48

When you are given sum and products of roots it is given by:

x2 – (Sum of roots)x + (product of roots) = 0

⇒      x2 – 14x + 48 =0

Hence, x2 – 14x + 48 =0 is the correct equation.

Example 2: A and B solved a quadratic equation in solving it, A made a mistake in the constant term and obtained the roots as 5,-3 while B made a mistake in the coefficient of x and obtained the roots as 1,-3. The correct roots of the equation are?

Sol: Let ax2+bx+c=0 is the correct equation
and ax2+bx+c1=0 is the first wrong equation hence b/a = -2
and ax2+b1x+c=0 is the second wrong equation c/a = -3 , there for a:b:c= 1:-2:-3

Therefore, the equation is x2 -2x -3, therefore on solving the roots are  3, -1

There is not that much variation in these type of questions, a basic concept practice would be enough.

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