Question

If ratio of the zeros of polynomial x×x-kx+6 is 3:2 the find the value of k

Answer 1

Answer:

k = 5

Step-by-step explanation:

Let the zeroes be α and β.

Given that they are in the ratio of 3 : 2.

Therefore,

α = 3x, β = 2x

Now, p(x) = x² - kx + 6 (given)

On comparing the above equation with ax² + bx + c, we get

a = 1, b = - k, c = 6

Sum of zeroes = α + β

= 3x + 2x

= 5x ...(i)

Also, sum of zeroes = - b/a

= - (- k)/1

= k ...(ii)

From (i) and (ii), we get

k = 5x ...(iii)

Product of zeroes = αβ

= (3x)(2x)

= 6x² ...(iv)

Also, product of zeroes = c/a

= 6/1

= 6 ...(v)

From (iv) and (v), we get

6x² = 6

On further solving, we get

→ x² = 6/6

→ x² = 1

→ x = √1

→ x = 1

Putting this value in (iii), we get

→ k = 5(1)

→ k = 5

Answer 2

$\bf{\Huge{\boxed{\rm{\pink{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

If the ratio of the zeroes of polynomial x² - kx + 6 is 3:2.

$\bf{\Large{\underline{\bf{To\:\:find\::}}}}$

The value of k.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

Given that ratio,

$\bf\begin{cases}\bf{\frac{\alpha }{\beta }\:=\:\frac{3}{2}\end{cases}}$

Compare x²-kx+6 with standard form of quadratic equation Ax² + Bx+C

• A = 1
• B = -k
• C = 6

A/q

Suppose α = 3R & β = 2R.

$\bf{\Large{\boxed{\sf{Sum\:of\:the\:zeroes\::}}}}$

$\longmapsto\sf{\alpha \:+\:\beta \:=\:\frac{-b}{a} }$

$\longmapsto\sf{3R \:+\:2R \:=\:\frac{-(-k)}{1} }$

$\longmapsto\sf{5R\:=\:k....................(1)}$

$\bf{\Large{\boxed{\sf{Product\:of\:zeroes\::}}}}$

$\longmapsto\sf{\alpha \:*\:\beta \:=\:\frac{c}{a} }$

$\longmapsto\sf{3R\:*\:2R\:=\:\frac{6}{1} }$

$\longmapsto\sf{6R^{2} \:=\:6}$

$\longmapsto\sf{R^{2} \:=\:\cancel{\frac{6}{6} }}$

$\longmapsto\sf{R^{2} \:=\:1}$

$\longmapsto\sf{R\:=\:\sqrt{1} }$

$\longmapsto\sf{\red{R\:=\:1}}$

Putting the value of R in equation (1), we get;

$\longmapsto\sf{K\:=\:5(1)}$

$\longmapsto\sf{\red{K\:=\:5}}$

Thus,

$\bf{\Large{\boxed{\rm{The\:value\:of\:K\:is\:\:5}}}}}$