Question

answer this question


thevirat ❤️
Bhai tumare liey hi ki hu ​

Answer 1

Let the ratio be t

according to question

(x - 3t)(X - 2t) = 0

x^2- 5t X + 6t^2 = 0

given equation

x^2 - kX + 6 = 0!

so

6t^2 = 6

t = +1 and -1

on comparing with coffecient of X we get

k = 5t

on putting the value of t we get

so the value of k is -5 or 5

Verification

Take

k = -5

x^2 + 5x + 6 = 0

x^2 + 2x +3X + 6 = 0

(X + 2) (X +3) = 0

so

X = -2 and -3

Ratio of zero of polynomial is

2/3 or 3/2

Again

for t = 5

x^2 - 5x + 6 =0

x^2 - 3x - 2x + 6 =0

(X - 3)(X - 2) =0

X = 3 and 2

so

Ratio of 0 of polynomial is

3/2

Hence solutions is right

and the value of t is 5 and -5

Answer 2

Given :

• The polynomial x² - kx + 6 has zeroes in the ratio 3:2.

To Find :

• Value of k

Solution :

Let x be the common multiple of the given ratio, 3:2

•°• Zeroes are → 3x and 2x

Let's find the sum of the zeroes.

$\implies$ $\sf{Sum\:of\:zeroes\:=\:3x+2x}$

$\sf{Sum\:of\:zeroes\:=\:5x\:\:...(i)}$

Now, let's find the product of the zeroes.

$\implies$ $\sf{Product\:of\:zeroes\:=\:3x\:\times\:2x}$

$\sf{Product\:of\:zeroes\:=6x^2\:...(ii)}$

We can find the value of sum of zeroes and product of zero in the given quadratic polynomial using,

$\large{\boxed{\sf{\red{x^2\:-(\alpha\:+\:\beta)x\:+\:(\alpha\:\beta)}}}}$

So, comparing the given polynomial with the form x² - (α+β)x + (αβ),

we can clearly see that product of zeroes is 6.

From equation (ii),

$\implies$ $\sf{6x^2=6}$

$\implies$ $\sf{x^2\:=\dfrac{6}{6}}$

$\implies$ $\sf{x^2=1}$

$\implies$ $\sf{x=\sqrt{1}}$

$\implies$ $\sf{x=\:\pm1}$

We know that, whenever a number is in a square root we obtain a negative and a positive value.

$\implies$ $\sf{x=\:1\:\:or\:x=-1}$

$\large{\boxed{\sf{\red{Value\:of\:x\:=\:1\:or\:-1}}}}$

Now, block in the value of x in sum of zeroes i.e 5x.

When x = 1 :

$\implies$ $\sf{3x+2x=k}$

$\implies$ $\sf{5x=k}$

$\implies$ $\sf{5(1)=k}$

$\implies$ $\sf{5=k}$

When x = - 1 :

$\implies$ $\sf{3x+2x=k}$

$\implies$ $\sf{5x=k}$

$\implies$ $\sf{5(-1)=k}$

$\implies$ $\sf{-5=k}$

$\large{\boxed{\sf{\purple{Value\:of\:k\:=\:5\:\:or-5}}}}$