Question

14.Write a quadratic polynomial, sum of whose zeroes is 2 and product is -8.​

Answer 1

Given:-

• Sum of zeroes = 2
• Product of zeroes = -8

To find:-

A quadratic Polynomial

Assumption:-

Let sum of zeroes be $\alpha+\beta$ = 2

Let the product of zeroes be $\alpha\beta$ = -8

Solution:-

We know,

Quadratic Equation is always in the form of:-

$\sf{x^2 + (\alpha+\beta)x + \alpha\beta}$

Substituting the values,

$\sf{x^2 + 2\times x + (-8)}$

= $\sf{x^2 + 2x -8}$

Therefore the quadratic Polynomial is$\sf{ x^2 + 2x - 8}$

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Verification:-

Let us find out the zeroes of polynomial

x² + 2x - 8

= $\sf{x^2 + 2x - 8}$

By splitting the middle term,

$\sf{x^2 + 4x - 2x - 8}$

= $\sf{x(x+4) -2(x+4)}$

= $\sf{(x+4)(x-2)}$

Either,

$\sf{x+4 = 0}$

= $\sf{x = -4}$

Or,

$\sf{x-2 = 0}$

= $\sf{x = 2}$

Now Let us find the sum of product of zeroes,

$\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{-4+2 = \dfrac{-2}{1}}$

= $\sf{-2 = -2}$ [Verified]

$\sf{Product\:of\:zeroes = \dfrac{Constant\:Term}{Coefficient\:of\:x^2}}$

= $\sf{4\times (-2) = \dfrac{-8}{1}}$

= $\sf{-8 = -8}$ [Verified]

Therefore the Quadratic Equation is $\sf{\boxed{\sf{x^2 + 2x - 8}}}$ Hence Verified.

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Answer 2

Answer:

$\Huge \pink{ \overline{ \underline{ {Given}}}}$

$•Sum \: of \: zeroes =2</h3><h3> \\ •Product \: of \: zeroes=-8$

$\large \green{ \underline{To \: Find}}$

$A \: quadratic \: Polynomial$

$\large \purple{ \underline{Assumption}}$

$Let \: sum \: of \: zeroes \: be \: \alpha + \beta = 2$

$Let \: the \: product \: of \: zeroes \: be \: \alpha \beta = - 8$

$\huge \underline{Solution:-}$

$we \: know,$

$Quadratic \: Equation \: is \: always \: in \: \\ the \: form \: of$

${x}^{2} + ( \alpha + \beta )x + \alpha \beta$

$\underline{Substituting \: the \: value}$

${x}^{2} + 2 \times x + ( - 8)$

$= {x}^{2} + 2x - 8$

$\therefore \: the \: quadratic \\ Polynomials \\ = {x}^{2} + 2x - 8$

$\huge{ \underline{Verification}}$

$Let \: us \: find \: out \: the \: \\ zeroes \: of \: polynomial$

${x}^{2} + 2x - 8$

$= {x}^{2} + 2x - 8$

$\underline{By \: spliting \: the \: middle \: terms}$

${x}^{2} + 4x - 2x - 8$

$= x(x + 4) - 2(x + 4)$

$= (x + 4)(x - 2)$

Either,

$x + 4 = 0 \\ = x = - 4$

Or,

$x - 2 = 0 \\ x = 2$

Now let us find the sum of product of zeroes,

$Sum \: of \: zeroes = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$= - 4 + 2 = \frac{ - 2}{1}$

$= - 2 = - 2[Verified]$

$Product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

$= 4 \times ( - 2) = \frac{ - 8}{1}$

$Therefore \: the \: quadratic \: equation \: \\ is \: {x}^{2} + 2x - 8$

Hence Verified.