Question

Write the quadratic polynomial, whose sum of zeroes is - 3, sum of the squares of zeroes is 17?

Answer 1

Answer:

x^2 + 3x - 4

Step-by-step explanation:

Let a and b are the roots.

  According to the question:

         a + b = - 3

⇒ a^2 + b^2 = 17

⇒ a^2 + b^2 + 2ab - 2ab = 17

⇒ ( a + b )^2 - 2ab = 17

⇒ ( - 3 )^2 - 2ab = 17

⇒ 9 - 2ab = 17

⇒ 9 - 17 = 2ab

⇒ - 8 = 2ab

⇒ - 4 = ab = product of roots.

   Hence,

Req. pol. is x^2 - ( sum of roots )x + product of roots

⇒ x^2 - ( a + b )x + ab

⇒ x^2 - ( - 3 )x + ( - 4 )

⇒ x^2 + 3x - 4

Answer 2

${\underline{\sf{Answer:-}}}$

${\sf{Let \: the \: zeros \: of}} \\ {\sf{polynomial \: be \: \alpha \: and \: \beta }} \\ {\sf{Given \: that :-}}\\ {\sf{\alpha + \beta = - 3}}$

${\sf{ {\alpha }^{2} + { \beta }^{2} = 17}}$

${\sf{( \alpha + \beta ) {}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta }} \\ {\sf{ {( - 3)}^{2} = 17 + 2 \alpha \beta }} \\ {\sf{9 = 17 + 2 \alpha \beta }} \\ {\sf{2 \alpha \beta = 9 - 17}} \\ {\sf{2 \alpha \beta = - 8}} \\ {\sf{\alpha \beta = \frac{ - 8}{2} }} \\ {\sf{\alpha \beta = - 4}}$

${\sf{In \: a \: quadratic \: equation:-}} \\ {\sf{ax {}^{2} + bx + c = 0}} \\ {\sf{\boxed{The \: sum \: of \: zeros = \frac{ - b}{a} }}} \\ {\sf{\boxed{The \: product \: of \: zeros = \frac{ c}{a} }}}$

${\sf{Therefore:- }} \\ {\sf{ \alpha + \beta = \frac{ - b}{a} ~ and \: \alpha \beta = \frac{c}{a} }} \\ {\sf{- 3 = \frac{ - b}{a} \: and \: - 4 = \frac{c}{a} }}\\ {\therefore}~{\sf{a=1}}$

Therefore :-

${\sf{a=1, b=3 ~and~ c=-4}}$

The equation is:-

$\red{\sf{{x}^{2} + 3x - 4}}$