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 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

Answer 2

$\large\green{\tt{\underline{\underline{Given:-}}}}$

${kx}^{2} - 14x + 8 = 0$

$\orange{\bold{\underline{\underline{Step - by - step - explanation:-}}}}$

$let \: the \: roots \: be \: \alpha and \beta$

A\Q,

$\alpha = 6 \beta$

Now,

$\alpha and \beta are \: the \: equation \: will \: be \: (x - \alpha ) \: and \: (x - \beta )$

$(x - \alpha ) (x - \beta )$

${x}^{2} - (\alpha + \beta)x + \alpha \beta = 0$

Now putting the value,

$\alpha = 6 \beta$

${x}^{2} -( 6 \alpha + \beta )x + 6 \alpha \times \beta = 0$

${x }^{2} - 7 \beta x + {6 \alpha }^{2} = 0$

Now, comparing with,

Now, comparing with

${k}^{2} - 14x + 8 = 0$

$7 \beta = \frac{14}{k}$

$\beta = \frac{2}{k}$

${ \beta }^{2} = \frac{4}{{k}^{2} } equation(1)$

$6 { \beta }^{2} = \frac{8}{k}$

${ \beta }^{2} = \frac{4}{3k} \: equation(2)$

From equation(1) and (2), we get....

$\frac{4}{k}^{2} \frac{4}{3k}$

$\beta = \frac{2}{k}$

${ \beta }^{2} = \frac{4}{ {k}^{2} }$

$\frac{4}{ {k}^{2} } = \frac{4}{3k}$

∴ k = 3.