Question

If alpha and beta, are the roots of the quadratic equation x2 + x + 1 = 0, then alpha/beta +beta/alpha =?​

Answer 1

Solution

Given :-

• Polynomial equation, x² + x + 1 = 0,
• roots are alpha & beta .

Find :-

• Value of alpha/beta + beta/alpha

Explanation,

let here,

• alpha = p
• beta = q.

So, Using Formula

★ Sum of roots = -(coefficient of x²)/(coefficient of x)

★ product of roots = (constant part)/(coefficient of x)

Then, now

==> Sum of roots = -1/1

==> p + q = -1___________(1)

Again,

==> Product of roots = 1/1

==> p .q = 1_____________(2)

By equ(1)

==> p = 1/q ___________(3)

Keep in equ(1)

==> q + 1/q = -1

==> q² + 1 + q = 0

Using Dharmacharya Formula

★ x = [-b ± √(b² - 4ab)]/2a

Where,

• a = 1
• b = 1
• c = 1.

keep value,

==> x = [ -1 ± √(1² - 4×1×1)]/2×1

==> x = [-1 ± √(1 - 4)]/2

==> x = [ -1 ± √(-3)]/2

we know,

• √-1 = i ( iota ) [ Imaginary Number ].

Show, keep

==> x = (-1 ± 3i)/2

First take (-ve) sign.

==> x = (-1 - 3i)/2

Now, take (+ve)sign .

==> x = (-1 + 3i)/2

Since,

• Roots be of this Equation ,p = (-1+3i)/2
• q = (-1 - 3i)/2

Now, Calculate value of p/q + q/p.

= p/q + q/p

= (p² + q²)/pq

keep value of p & q

= [ {-1+3i)/2}² + {(-1-3i)2}²]/[(-1+3i)/2 × (-1-3i)/2]

= [ 1 + 9i² - 6i + 1 + 9i² + 6i)/4]/[-(1 - 9i²)/4]

= [(2 + 18i² )/4] × 4/ (9i² - 1)

We know,

• i² = -1

Then

= [ (2 - 18)/4 × 4/(-9-1)

= -16/4 × -4/10

= -4 × -2/5

= 8/5 [Ans]

Hence

• Value will be (alpha/beta + beta/alpha) = 8/5 .

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