Question

If one root of the quadratic equation kx2-14x+8=0 is 6 times the other, then find the value of k.

Answer 1

Answer:

Let One Root be n and Other Root be 6n.

⋆ Given Polynomial : kx² – 14x + 8 = 0

$\underline{\bigstar\:\textbf{Sum of Zeroes :}}$

$:\implies\sf Sum\:of\:Zeroes=-\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^2}\\\\\\:\implies\sf n+6n=\dfrac{-\:(-14)}{k}\\\\\\:\implies\sf 7n = \dfrac{14}{k}\\\\\\:\implies\sf n = \dfrac{2}{k} \qquad \dfrac{ \quad}{}\:eq.(1)$

$\rule{130}{1}$

$\underline{\bigstar\:\textbf{Product of Zeroes :}}$

$:\implies\sf Product\:of\:Zeroes=\dfrac{Constant\:Term}{Coefficient\:of\:x^2}\\\\\\:\implies\sf n \times 6n=\dfrac{8}{k}\\\\\\:\implies\sf 6n^2=\dfrac{8}{k}\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting value of n from eq. (1)}}\\\\:\implies\sf 6 \times \left(\dfrac{2}{k}\right)^2=\dfrac{8}{k}\\\\\\:\implies\sf 6 \times \dfrac{4}{k^2} = \dfrac{8}{k}\\\\\\:\implies\underline{\boxed{\sf k = 3}}$

⠀

$\therefore\:\underline{\textsf{Hence, Value of k will be \textbf{3}}}.$

Answer 2

Answer:

Let roots be α and β

A/q

α = 6β

now, if α and β are roots then equation will be (x -α)(x -β) =0

(x -α)(x -β) =0

⇒ x² - (α+β)x + αβ =0

now putting α = 6β ,

⇒x² - (6β +β)x + 6β×β =0

⇒x² - 7βx +6β² =0

now comparing with kx² -14x +8 =0

7β =14/k

⇒β =2/k

⇒β² = 4/k²_______(1)

and 6β² =8/k

⇒β² =4/3k_______(2)

equating (1) and (2), we get,

4/k² = 4/3k

⇒k =3