Question

If zeroes of polynomial 6x² + 13x + 2a are t and 1/t
then find the value of a a and also
the zoroes of polynomial​

Answer 1

Answer:

Step-by-step explanation:

sum of roots t+1/t=-13/6---(1)

product of roots=2a/6=a/3

t*1/t=a/3

a=3

now

(t+1/t)^2=169/36

t^2+1/t^2+2=169/36

t^2+1/t^2=169/36-2

t^2+1/t^2=169-72/36=97/36

(t-1/t)^2=t^2+1/t^2-2=97/36-2=97-72/36=25/36

t-1/t=5/6---(2)

(1)+(2)

2t=-8/6

t=-2/3

1/t=-3/2

Answer 2

Answer :

The value of a is 3

and the zeroes are -3/2 and -2/3

Given :

The quadratic polynomial is :

• 6x² + 13x + 2a
• The zeroes of the polynomial are t and 1/t

To Find :

The value of a and the zeroes of the polynomial.

Solution :

We know that

$\sf{sum \: of \: the \: zeroes = \dfrac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }$

and

$\sf{product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } }$

So taking the Product of the zeroes in the given case :

$\sf{t \times \dfrac{1}{t} = \dfrac{2a}{6} } \\ \\ \implies \sf1 = \dfrac{a}{3} \\ \\ \implies \sf{a = 3}$

The value of a is 3

Putting the value of a the quadratic polynomial we obtain :

$\sf{6 {x}^{2} + 13x + 2(3) } \\ \sf= 6 {x}^{2} + 13x + 6$

Now Factorizing the polynomial we have :

$\sf6 {x}^{2} + 13x + 6 \\ \sf = 6 {x}^{2} + 9x + 4x + 6 \\ \sf = 3x(2x +3 ) + 2(2x + 3) \\ \sf = (2x + 3)(3x + 2)$

So the zeroes are :

$\sf 2x + 3 = 0 \: \: and \: \: 3x + 2 = 0 \\ \sf \implies x = - \frac{3}{2} \: \: and \implies x = - \frac{ 2}{3}$