Question

write a quadratic polynomial sum of whose zeroes is -3 and product is -7​

Answer 1

Answer:

${x}^{2} + 3x - 7$

Answer 2

$\large{\boxed{\red{\tt Given :-}}}$

• Sum of zeroes of quadratic polynomial is -3 .
• Product of zeroes of quadratic polynomial is -7

$\large{\boxed{\purple{\tt To\:Find-}}}$

• The quadratic polynomial .

$\large{\boxed{\pink{\tt Concept\:Used :-}}}$

Zeroes of a quadratic polynomial in standard form ax²+bx +c is given by the Quadratic formula. Also , sum of zeroes equals to $\sf\frac{-b}{a}$ and the Product of zeroes is given by $\sf\frac{c}{a}$ . So , if the sum and Product of zeroes of is given we can find the quadratic polynomial as , x² - ( α + β )x + α β , where α & β are the zeroes of quadratic polynomial.

$\large{\boxed{\green{\tt Answer :-}}}$

Now ,here ,

• Sum of zeroes = -3
• Product of zeroes = -7 .

On substituting the respective values in the formula stated ,

$\tt:\implies p(x)=k[x^2-(\alpha+\beta)+\alpha\beta \\\\\tt:\implies p(x)=k[x^2-( -3)x+(-7)\\\\\underline{\boxed{\orange{\bf\longmapsto p(x)=k[x^2+3x-7]}}}$

$\rule{200}2$

$\large{\boxed{\red{\tt Derivation\:of\: Formula\:used :-}}}$

If we consider a quadratic equation in standard form of ax² + bx + c = 0 . And let us suppose that α & β are its roots . So , the quadratic polynomial will be ,

$\tt:\implies (x-\alpha)(x-\beta)=0$

$\tt:\implies x(x-\beta)-\alpha(x-\beta)=0$

$\tt:\implies x^2-x\beta-x\alpha+\alpha\beta=0$

$\tt:\implies x^2-x(\alpha+\beta)+\alpha\beta=0$

$\underline{\boxed{\color{green}{\leadsto \sf p(x)=k[x^2-(\alpha+\beta)x+\alpha\beta]}}}$