Question

find the value of k for which one root of the quadratic equation kx2-14x+8=0 is six times the other

Answer 1

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

Answer:

k = 3

Step-by-step explanation:

Let the 2 roots of the equation be α and β

According to the given condition, α = 6β

Therefore (x-α) (x-β) = 0

x² - (α+β) + αβ = 0

Substituting α = 6β ,

x² - (6β+β) + 6β×β = 0

x² - 7β + 6β² = 0  

Now, comparing with  kx² -14x +8 =0

7β =14/k

β =2/k

β² = 4/k² ------------ (1)

On comparing we also get that ,

6β² =8/k

β² =4/3k -------------- (2)

From (1) and (2) ,

4/k² = 4/3 k

∴k = 3