Question

Find the quadratic polynomial
whose zeroes are 2 and - 3
verify the relation b/w the
cofficients, and the zeroes of
the polynomial

​

Answer 1

Answer:

${x}^{2} + x - 6$

Step-by-step explanation:

let the two zeroes be alpha and beta respectively

$\alpha = 2 \\ \beta = - 3 \\ \alpha + \beta = \frac{ - b}{a} \\ \alpha \times \beta = \frac{c}{a}$

thus, according to the above identities,

$2 + ( - 3) = \frac{ - b}{a} \\ \frac{ - b}{a} = - 1 \\ - - - \times - - - \\ 2 \times - 3 = \frac{c}{a} \\ \frac{c}{a} = - 6$

thus, from the above given two values,

$a = 1 \\ b = 1 \\ c = - 6$

thus, the required quadratic equation is :

${x}^{2} + x - 6$

VERIFICATION:

$\alpha = 2 \: \: \: \: \: \beta = - 3 \\ \alpha + \beta = 2 + ( -3) \\ = 2 - 3 = - 1 \\ \frac{ - b}{a} = \frac{ - 1}{1} = - 1$

Similarly ,

$\alpha \times \beta = 2 \times - 3 \\ = - 6 \\ \frac{c}{a} = \frac{ - 6}{1} = - 6$

Hence , proved the relationship.

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