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

1
by akkkkk1

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by akkkkk1

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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**

A/q

(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

⇒