Question

Integral zeroes of the polynomial ( x + 3)(x-7) are
a) -3, -7
b) 3,7
C) -3,7


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Answer 1

\begin{lgathered}\alpha = 7 + \sqrt{3} \\ \beta = 7 - \sqrt{3}\end{lgathered}

α=7+

3

β=7−

3

We know that,

\begin{lgathered}\alpha + \beta = \frac{ - b}{a} \\ \alpha \beta = \frac{c}{a}\end{lgathered}

α+β=

a

−b

αβ=

a

c

So,

b = (- 49)

a = 1

c = 46.

By general Quadratic equation,

a {x}^{2} + bx + c = 0ax

2

+bx+c=0

Quadratic equation is:-

{x}^{2} - 49x + 46x

2

−49x+46

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Answer 2

Answer:

option C is correct

Step-by-step explanation:

bcoz here x+3. x-7

are zeroes of the polynomial

so

(x+3)(x-7).= 0

now

x+3=0

x= -3

x-7=0

x=7