Question

Q7. If a and b are the zeroes of the
polynomial x2 -6x +k. Find the value of
k, if a +b = 12ab​

Answer 1

Answer:

k= 1/2

Step-by-step explanation:

a+b= 6

ab=k

12ab=a+b

12ab=6

ab=1/2

k=1/2

Hope this helps.

Answer 2

$\bf{\underline{\underline{Given}}}$

$\mathsf{\star\:\: Polynomial = x^{2} - 6x + k}$

$\mathsf{\star\:\: a\:\: and\:\:b\:\:are\:the\:zeroes}$

$\mathsf{\star\: \:a + b = 12ab}$

$\bf{\underline{\underline{To\:find}}}$

$\mathsf{\star\: Value\:of\:k}$

$\bf{\underline{\underline{Solution}}}$

$\mathtt{a + b = 12ab\:\:\: \rightarrow (1) \: [Given]}$

$\text{\underline{We know that}}$

$\mathsf{Sum\:of\:zeroes = \frac{-(x\:coefficient) }{x^{2}\: coefficient}}$

$\mathtt{a + b = \frac{-(-6)}{1}\: }$

$\mathtt{\implies \: a + b = 6 \:\:\: \rightarrow (2)}$

From (1) and (2) :

$\mathtt{\implies\: 12ab = 6}$

$\mathtt{\implies\: ab = \frac{6}{12}}$

$\mathtt{\implies\: ab = \frac{1}{2}}$

$\text{\underline{We know that}}$

$\mathsf{Product\:of\:zeroes = \frac{constant}{x^{2}\:coefficient}}$

$\mathtt{\implies\: ab = \frac{k}{1}}$

$\mathtt{\implies\: ab = k}$

$\fbox{\mathtt{\green{\implies\: \frac{1}{2} = k}}}$

∴ Value of k = 1/2