Question

find the quadratic polynomial whose zeros are 3 and -4

Answer 1

Answer:

The obtained polynomial is $x^2+x-12$ having zeroes as 3 and -4.

To find:

Quadratic polynomial

Solution:

Given : Zeros of quadratic polynomial are 3 and -4.

Sum of zeros = 3 + (-4)  = -1

Product of zeros = (3)(-4) = -12

Now the required polynomial is  

$\begin{array} { c } { x ^ { 2 } + ( \text {product of zeros} ) - x ( \text {sum of zeros} ) } \\\\ { = x ^ { 2 } + ( - 12 ) - x ( - 1 ) } \\\\ { = x ^ { 2 } - 12 + x } \\\\ { = x ^ { 2 } + x - 12 } \end{array}$

Thus, the obtained polynomial is $x^2+x-12$ having zeroes as 3 and -4.

Answer 2

Solution:

Let the quadratic polynomial be ax²+bx+c, a≠0 and its zeroes be $\alpha \: and \beta,$

Here , $\alpha = 3 , \beta = -4$

i ) Sum of the zeroes

= $\alpha +\beta$

= $3+(-4)$

=$-1$

$\alpha +\beta=-1$---(1)

ii) product of the zeroes

= $\alpha\beta$

=$3\times(-4)$

=$-12$

$\alpha\beta=-12$--(2)

Therefore,

the quadratic polynomial ax²+bx+c is

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

, where k is a constant

= $k[x^{2}-(-1)x+(-12)$

/* From (1) and (2) */

= $k(x^{2}+x-12)$

we can put different values of k.

When k=1,

the quadratic polynomial will be x²+x-12

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