Question

if alfa and beta are zero's of polynomial,p(x)=6x²-5x+k such that alfa -beta=1/b , find the value of k​

Answer 1

Answer:

answer for the given problem is given

Answer 2

${\huge{\sf {\orange {\underline{Question :}}}}}$

$\: \:$

If α and β are zeroes of polynomial

$p(x) = 6 {x}^{2} - 5x + k$

such that

$\alpha - \beta = \frac{1}{a}$

find the value of k.

$\: \:$

${\huge{\sf {\green{\underline{Answer}}}}}$

$\: \:$

$\boxed{ k \:\: = \:\: 1 }$

$\: \:$

${\huge{\sf {\blue{\underline{Explanation : }}}}}$

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Formula used :

$\: \:$

• $sum \: of \: zeroes \: = \: \frac{ - b}{a}$
• $product \: of \: zeroes \: = \: \frac{c}{a}$
• ${(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$\: \:$

Solution :

$\: \:$

$sum \: of \: zeroes \: = \: \frac{ - ( - 5)}{6}$

$\alpha + \beta = \frac{5}{6}$

$\: \: \:$

$product \: of \: zeroes \: = \: \frac{k}{6}$

$\alpha \times \beta = \frac{k}{6}$

$\: \: \:$

$( \alpha - \beta ) = \frac{1}{a}$

$( \alpha - \beta ) = \frac{1}{ 6}$

$\: \:$

${(a + b)}^{2} - {(a - b)}^{2} = 4ab$

${( \alpha + \beta )}^{2} - {( \alpha - \beta )}^{2} = 4 \alpha \beta$

${ (\frac{5}{6} )}^{2} -{ ( \frac{1}{6} )}^{2} = 4 \alpha \beta$

$4 \alpha \beta = \frac{25}{36} - \frac{1}{36}$

$4 \alpha \beta = \frac{24}{36}$

$\alpha \beta = \frac{24}{36 \times 4}$

$\alpha \beta = \frac{6}{36}$

$\alpha \beta = \frac{1}{6}$

$\: \: \:$

$\frac{k}{6} = \alpha \beta$

$\frac{k}{6} = \frac{1}{6}$

$k = \frac{1 \times 6}{6}$

$k = 1$