Question

If one zero of (k+1)^2-5x+5 is multiplicative inverse of the other, find zeroes of kx^2 + 3x+9, where k is constant.​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

Let the roots of the Q.E $\sf{(k + 1) {x}^{2} - 5x + 5 \: \: be \: \alpha \: and \: \beta \: respectively}$

As roots are multiplicative inverse of one another,

$\sf{ \therefore \: let \: \beta = \frac{1}{ \alpha } }$

Now,

$\sf{Product \: of \: zeroes = \frac{c}{a}}$

$\sf {\implies \: \alpha \times \beta = \frac{c}{a}}$

$\sf {\implies \: \alpha \times \frac{1}{ \alpha } = \frac{5}{k + 1}}$

$\sf {\implies \: 1 = \frac{5}{k + 1}}$

$\implies \: \sf{k + 1 = 5}$

$\implies \: \boxed{\sf{ k = 4}}$

Now,

$\implies \sf{k {x}^{2} - 3kx + 9 = 0}$

Putting the value of 'k'

$\implies \sf{4{x}^{2} - 3(4)x + 9 = 0}$

$\implies \sf{4{x}^{2} - 12x + 9 = 0}$

$\implies \sf{4{x}^{2} - 6x - 6x + 9 = 0}$

$\sf{ \implies \: 2x(2x - 3) - 3(2x - 3) = 0}$

$\sf{ \implies \:( 2x - 3)(2x - 3) = 0}$

$\sf{ \implies \: x = \frac{3}{2} \: or \: x = \frac{3}{2}}$

$\boxed{\: \sf{ Hence, \: the \: required \: zeroes \: are \: \frac{3}{2} and \: \frac{3}{2} }}$

Answer 2

Answer:

2/3

Step-by-step explanation:

if zeros of the polynomial (k+1)x^2-5x+5 is the multiplicative inverse of the other then product of zeros is 1 hence 5/(k+1)=1 then k=4 now,kx^2-3kx+9 becomes 4x^2-12x+9 which is the square of (2x-3)^2 hence the zero is 2/3