Question

If sum and product of zeroes of quadratic
polynomial are, respectively 8 and 12, then find
their zeroes.
​

Answer 1

Answer:

required quadratic polynomial is \({x^2} - 8x + 12\) and zeroes are 2 and 6.

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Answer 2

● Given :-

• Sum of the Zeroes = 8
• Product of Zeroes = 12

● To find :-

• Their zeroes.

● Solution :-

Here we have given Sum and Product of polynomial as 8 & 12. First we have to find Quadratic Polynomial and then Zeroes. Also wr know that Quadratic Polynomial is always in the form of ax² + bx + c where a, b and c are real number.

▩ Method No.1

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✬ Quadratic Polynomial is

✹ Sum of the Zeroes :-

${\bold{\implies{ \alpha + \beta = \cfrac{ - b}{a} }}} \\ \\ {\bold{\implies{ \alpha + \beta = \cfrac{ - 8}{1} }}}$

✹ Now, Product of the Zeroes :-

${\bold{\implies{ \alpha \beta = \dfrac{c}{a} }}} \\ \\ {\bold{\implies{ \alpha \beta = \dfrac{12}{1} }}}$

✹ Form here we have :-

• a = 1
• b = -8
• c = 12

✹ Quadratic Polynomial form is :-

${\bold{\implies{a {x}^{2} + bx + c}}} \\{\bold{\implies{(1){x}^{2} + ( - 8)x + (12)}}} \\ \blue{\bold{\implies{ {x}^{2} - 8x + 12}}}$

✹ Factorization of Polynomial :-

${\sf{\longmapsto{ {x}^{2} - 8x + 12 = 0}}} \\ {\sf{\longmapsto{ {x}^{2} - 6x - 2x + 12 = 0}}} \\ {\sf{\longmapsto{ x(x - 6) - 2(x - 6) = 0 }}} \\ \blue{\sf{\longmapsto{ (x - 2)(x - 6) = 0}}}$

✬ Now, Zeroes of the Polynomial is :-

${\large{\sf{\longmapsto{x-2 =0}}}} \\ {\large{ \underline{ \boxed{ \red{\sf{\longmapsto{x =2}}}}}}}$

And

${\large{\sf{\longmapsto{x-6=0}}}} \\ {\large{ \underline{ \boxed{ \red{\sf{\longmapsto{x =6}}}}}}}$

So, 2 and 6 are the Zeroes of the polynomial.

_________________

▩ Method No. 2

✹ Required quadratic polynomial is

= x² - (α + β)x + αβ

= x² - 8x + 12

✹ Factorization of Polynomial :-

= x² - 8x + 12

= x² - 6x - 2x + 12

= x(x - 6) - 2(x - 6)

= (x - 2 ) (x - 6)

✬ Now, Zeroes of the Polynomial is :-

》 x - 2 = 0

》 x = 2

and

》 x - 6 = 0

》 x = 6

So, 2 and 6 are the Zeroes of the Polynomial.