Question

If a and b are the zeroes of p(x)=6x^2-7x+2. Find the quadratic polynomial whose zeroes are 1/a amd1/b.

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Answer 1

Step-by-step explanation:

given polynomial is

$6 {x}^{2} - 7x + 2$

sum of roots

$a + b = - ( - 7) \div 6$

$= \frac{ - 7}{6}$

product of roots

$ab = \frac{2}{6}$

the new quadratic equation with zero's

$\frac{1}{a } and \: \: \frac{1}{b}$

sum of roots

$\frac{1}{a} + \frac{1}{b}$

$= \frac{a + b}{ab}$

$= \frac{ \frac{ - 7}{6} }{ \frac{2}{6} }$

$= \frac{ -7 }{2}$

product of roots

$\frac{1}{a} \frac{1}{b}$

$= \frac{1}{ab}$

$= \frac{1}{ \frac{2}{6} }$

$= \frac{6}{2} = 3$

the quadratic polynomial with roots 1/a, 1/b is

${x}^{2} - ( \frac{1}{a} + \frac{1}{b} )x + \frac{1}{a} \times \frac{1}{b}$

${x}^{2} - ( \frac{ - 7}{2} )x + 3$

${x}^{2} + \frac{7}{2} x + 3$

$\frac{1}{2} (2 {x}^{2} + 7x + 6)$

Answer 2

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