Question

Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively
(v) -1/4,1/4​

Answer 1

${\underline{\underline{\bold{Question:}}}}$

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively $- \dfrac{ 1}{4}$, $\dfrac{ 1}{4}$.

${\underline{\underline{\bold{Answer :}}}}$

${\underline{\sf{Given:}}}$

• Sum of the Zeroes = $- \dfrac{ 1}{4}$

• Product of the Zeroes = $\dfrac{ 1}{4}$

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

• We have to find Quadratic Polynomial.

${\underline{\sf{Solution:}}}$

● We know that Quadratic Polynomial is always in the form of ax²+ bx + c. Also

${\underline{\boxed{\sf{Sum \: of \: the \: Zeroes = \alpha + \beta = \dfrac{ - b}{a} }}}}$

and

${\underline{\boxed{\sf{Product \: of \: the \: Zeroes = \alpha \beta = \dfrac{ c}{a} }}}}$

● So from here, value of b & a is,

${\longmapsto{\sf{ \alpha + \beta = \dfrac{-b}{a} = - \bigg( \cfrac{ - 1}{4} \bigg) }}} \\ {\longmapsto{\sf{ \dfrac{ - b}{a} = \frac{ 1}{4} }}}$

Here

• a = 4
• b = 1

● Second, value of c is

${\longmapsto{ \sf{ \alpha \beta = \dfrac{c}{a} = \bigg( \dfrac{1}{4} \bigg) }}}$

Here

• c = 1
• a = 4

● Now we have

• a = 4
• b = 1
• c = 1

● Substituting the values in the standard form of quadratic polynomial :-

${\large{\sf{\implies{a {x}^{2} + bx + c}}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{\sf{\implies{(4){x}^{2} + (1)x + (1)}}}} \\ { \underline{ \boxed{\large{ \red{\sf{\implies{4 {x}^{2} + x + 1}}}}}}}$

So, here Quadratic polynomial is 4x² + x + 1 .