Question

if alpha and beta are multiple zeroes of 6x^2-5x-k and alpha-beta=7/6 so find the value of k​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{QUESTION:-}}}}}}$.

• if $\alpha\:and\:\beta$ are multiple zeroes of $\:(6x^2-5x-k)$ and $\alpha\:-\:\beta\:=\:\frac{7}{6}$so find the value of k

$\large{\underline{\underline{\mathfrak{\green{\sf{Sulution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• Equation is $\:(6x^2-5x-k)\:=\:0$

• $\alpha\:and\:\beta\:are\:zeros\:of\:this\:equation$

• $\alpha\:-\:\beta\:=\frac{7}{6}$...(1)

$\large{\underline{\underline{\mathfrak{\sf{Find\:Here:-}}}}}$.

• Value of k

$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

We know ,

$\bold{\boxed{\boxed{\:Sum\:of\:Zeros\:=\frac{-(Coefficient\:of\:x)}{(Coefficient\:of\:x^2)}}}}$

So,

$\implies\:(\alpha\:+\beta)\:=\frac{-(-5)}{6}$

$\implies\:(\alpha\:+\beta)\:=\frac{5}{6}$...(2)

Addition of equation (1) and (2) ,

We find here,

$\implies\:2\alpha\:=\frac{7}{6}\:+\frac{5}{6}$

$\implies\:2\alpha\:=\frac{(7+5)}{6}$

$\implies\:2\alpha\:=\frac{\cancel{12}}{\cancel{6}}$

$\bold{\boxed{\boxed{\:\alpha\:=\:1}}}$

Keep value of $\alpha\:=\:1$ in (2)

we find here,

$\implies\:(1\:+\beta)\:=\frac{5}{6}$

$\implies\:(\beta)\:=\frac{5}{6}\:-\:1$

$\implies\:(\beta)\:=\frac{(5-6)}{6}$

$\bold{\boxed{\boxed{\:(\beta)\:=\frac{-1}{6}}}}$

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Again, We know ,

$\bold{\boxed{\boxed{\:Product\:of\:Zeros\:=\frac{(Constant\:part)}{(Coefficient\:of\:x^2)}}}}$

$\implies\:(\alpha\times\beta)\:=\frac{(-k)}{6}$

Keep value of $\alpha\:and\:\beta$

$\implies\:(1\:\times\frac{-1}{6})\:=\frac{k}{6}$

$\implies\:\frac{k}{6}\:=\frac{-1}{6}$

$\implies\:(k)\:=\frac{\cancel{-6}}{\cancel{6}}$

$\bold{\boxed{\boxed{\:(k)\:=\:-1}}}$

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$\large{\underline{\underline{\mathfrak{\green{\sf{Answer:-}}}}}}$.

• Value of k = -1

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