Question

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(i) Find a quadratic polynomial with zeroes -2 and -1/3.
(ii) What is the quadratic polynomial whose sum of the zeroes is -3/2.:and the product
of the zeroes is -1.​

Answer 1

Answer:

Step-by-step explanation:

1. the zeros of the polynomial are -2,-1/3

(x-2)(x-1/3)

x^2-x/3-2x+2/3

3x^2-x-6x+6=0

3x^2-7x+6=0

2. sum of xeros is -3/2

product of zeros is -1

let the polynomial be k(x^2-(sum of zeros)x+product of zeros)

k(x^2-(-3/2)x+(-1))

k(2x^2+3x-2)

2x^2+3x-2 is our required polynomia;

Answer 2

i) Given:

• Zeroes of the polynomial are -2 and -1/3

To find:

• The polynomial

Solution:

Let the zeroes be a and b

$\texttt{Sum of zeroes} = a + b\\ = > - 2 + ( - \frac{1}{3} ) \\ \\ = > - 2 - \frac{1}{3} \\ \\ = > \frac{ - 6 - 1}{3} = \frac{ - 7}{ \: 3} \\ \\ and \\ \texttt{ Product of zeroes} = ab \\ = > - 2 \times ( - \frac{1}{3} ) \\ \\ = > \frac{2}{3}$

Required polynomial =

$\blue{{ {x}^{2} - ( {a + b})x + ( { ab})}}$

Putting values

k is constant value here

$k \left( {x}^{2} - (\frac{ - 7}{ \: 3} )x + \frac{2}{3} \right) \\ \\ \rightarrow \: k \left( {x}^{2} + \frac{7}{3} x + \frac{2}{3} \right) \\ \\ \rightarrow \: k \left( \frac{ {3x}^{2} + 7x + 2 }{3} \right) \\ \\ \implies \: \frac{k}{3} \left( {3x}^{2} + 7x + 2 \right)$

Hence, the Required polynomial is

$\green{ {3x}^{2} + 7x + 2 }$

Another method is by making factors

so,

x = -2 and x = -1/3

[x-(-2) ] and [x-(-1/3)]

=> (x+2) and (x+1/3)

$=> (x+2)(x+\frac{1}{3}) =0$

$=> x(x+\frac{1}{3}) +2(x +\frac{1}{3}) =0$

$=> {x}^{2}+ \frac{x}{3 }+2x + \frac{2}{3} =0$

$=>\frac{{3x}^2 +x +6x +2 }{3 }=0$

$=>\green{ {3x}^{2} + 7x + 2 }$

__________________________

ii) Given:

• Sum of zeroes = -3/2
• Product of zeroes = -1

To find:

• The polynomial

Solution:

We know that

Polynomial is given by

${x}^{2} - ( Sum \: of \: zeroes)x + (Product \: of \: zeroes)$

K is just any constant

So,

The required polynomial =

${x}^{2} - ( - \frac{3}{2} ) x+ ( - 1) \\ \\ \implies \: k \left({x}^{2} + \frac{3}{2} x - 1 \right) \\ \\ \implies \: k \left ( \frac{ {2x}^{2} + 3x - 2 }{2} \right) \\ \\ \implies \frac{k}{2} \left ( {2x}^{2} + 3x - 2\right)$

Hence, the Required polynomial is

$\green{ {2x}^{2} + 3x - 2}$