Question

A quadratic polynomial, whose zeroes are –3 and 4, is
(A) x2
– x + 12 (B) x2
+ x + 12
(C)
2
– –6
2 2
x x
(D) 2x2
+ 2x –24

Answer 1

Question:

Find a quadratic polynomial, whose zeroes are –3 and 4.

Options:

(A). x^2 – x + 12

(B). x^2 + x + 12

(C). x^2 - x - 12

(D). 2x^2 + 2x –24

Answer:

Option (C).

x^2 - x - 12

Note:

If x = A and x = B are the zeros of given quadratic polynomial p(x) ,then p(x) will be given by;

p(x) = x^2 - (A+B)•x + A•B .

Solution:

Here,

The given zeros of required polynomial are

x = -3 and x = 4 .

Thus,

The required polynomial p(x) will be given by;

=> p(x) = x^2 - {(-3) + 4}•x + (-3)•4

=> p(x) = x^2 - x - 12

Hence,

The required polynomial is x^2 - x - 12 .

Answer 2

Question :-----

A quadratic polynomial, whose zeroes are –3 and 4, is

a) x²-x+12

b) x² - x - 12

c) x² - 6x + 12

d) 2x² + 2x - 24

we have to Find the correct Equation ?

$\Large\bold\star\underline{\underline\textbf{Concept\:used}}$

$\sf \: if \alpha \: and \: \beta \: are \: roots \: of \:quadratic\:polynomial\: \\ \sf \: than \: the \:polynomial\: is \: given \: by : - \\ \\ \bf \: {x}^{2} - (sum \: of \: roots)x + product \: of \: roots\\ or \\ \\ \red{\boxed{ \bf {x}^{2} - ( \alpha + \beta)x + \alpha \times \beta}}$

$\rule{200}{4}$

$\sf \: putting \: \alpha \: = - 3 \: and \: \beta = 4 \: we \: get \\ \\ \red\longrightarrow \: {x}^{2} - ( - 3 + 4)x + ( - 3 \times 4)\\ \\ \red\longrightarrow \: {x}^{2} - (1)x + ( - 12)\\ \\ \green{\boxed{\red\longrightarrow \: {x}^{2} - x - 12}}$

Hence, our Correct Answer is Option B ..

$\rule{200}{4}$

$\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

→ sum of Roots of Equation ax² + bx + c = 0 is given by = (-b/a) .

→ Product of Roots is given by = (c/a)

_______________________

$\large\underline\textbf{Hope it Helps You.}$

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