Question

8. Find a quadratic polynomial whose sum
Product of its
zeroes are 1 by 4 and -1
respectively
​

Answer 1

Answer:

$format \: : \\ let \: the \: requier \: valuable \: be \: x \\ {x}^{2} - (sum \: of \: root)x + (product \: of \: root) = 0 \\ {x}^{2} - ( \frac{1}{4} )x + ( - 1) = 0 \\ {x}^{2} - \frac{1}{4} x - 1 = 0 \\ \frac{ {x}^{2} }{1} - \frac{1}{4} x - \frac{1}{1} = \frac{0}{1} \\ take \: lcm \\ \frac{4 {x}^{2} - x - 4 }{4 } = 0 \\ 4 {x}^{2} - x - 4 = 0 \times 4 = 0$

Answer 2

Answer:

$Given, \: Sum \: of \: zeroes \: = 1/4</p><p> \\ and \: product \: of \: zeros = \: -1 \\ </p><p>Then, \: the \: quadratic \: polynomial \\ = k \: [ {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes ] \\ k( {x}^{2} - ( \frac{1}{4} )x - 1 \\ = k( {x}^{2} - \frac{x}{4} - 1) \\ = k( \frac{4 {x}^{2} - x - 4 }{4 } ) \\ If \: k = 4, \: then \: the \: required \: quadratic \: polynomial \: is \\⇒ \: 4 {x}^{2} - x - 4</p><p>$