Question

If m and n are zeroes of x²+5x-6.then write a quadratic polynomial whose zeroes are m/n & n/m.​

Answer 1

Answer:

6x² + 37x + 6

Step-by-step explanation:

For zeroes, polynomial = 0:

x² + 5x - 6 = 0 → x² + 6x - x - 6 = 0

=> x(x + 6) - (x - 1) = 0

=> (x + 6)(x - 1) = 0

=> x = - 6 & x = 1

thus, m and n are - 6 and 1.

m/n = -6/1 = - 6

n/m = -1/6

Required polynomial is:

=> (x - (-6)) (x - (-1/6)

=> (x + 6)(x + 1/6)

=> x² + x/6 + 6x + 1

=> 1/6 (6x² + x + 36x + 6) {Neglect numeric factor}

=> (6x² + 37x + 6) is the required polynomial.

Answer 2

Answer:

Let the m and n be the zeroes.

$\underline{\bigstar\:\textsf{Factorizing the Polynomial :}}$

$\dashrightarrow\sf x^2+5x-6=0\\\\\\\dashrightarrow\sf x^2+6x-x-6=0\\\\\\\dashrightarrow\sf x(x+6)-1(x+6)=0\\\\\\\dashrightarrow\sf (x-1)(x+6)=0\\\\\\\dashrightarrow\sf x=1\qquad or\qquad x=-\:6\\\\\\\dashrightarrow\sf m=1\qquad or\qquad n=-\:6$

$\rule{180}{1.5}$

Let the zeroes be $\sf\frac{m}{n}$ and $\sf\frac{n}{m}$

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf x^2-(Sum\:of\:Zeroes)x+Product\:of\:Zeroes = 0\\\\\\:\implies\sf x^2 -(\alpha + \beta)x + ( \alpha \beta) = 0\\\\\\:\implies\sf x^2 -\bigg(\dfrac{m}{n} + \dfrac{n}{m}\bigg)x +\bigg(\dfrac{m}{n} \times \dfrac{n}{m}\bigg) = 0\\\\\\:\implies\sf x^2 -\bigg(\dfrac{m^2 + n^2}{nm}\bigg) + 1 = 0\\\\\\:\implies\sf x^2 -\bigg(\dfrac{(1)^2 + ( - 6)^2}{1 \times - 6}\bigg) + 1 = 0\\\\\\:\implies\sf x^2 -\bigg(\dfrac{1 + 36}{-\:6}\bigg) + 1 = 0\\\\\\:\implies\sf x^2 + \dfrac{37}{6} + 1 = 0\\\\\\:\implies\underline{\boxed{\sf 6x^2 + 37x + 6 = 0}}$

⠀

$\therefore\:\underline{\textsf{Required quadratic equation is \textbf{6x ^\text2 + 37x + 6}}}.$