Question

zeroes of a polynomial are 3/2 & 2/3 then find the polynomial​

Answer 1

Answer:

6x² - 13x + 6

Step-by-step explanation:

It is given that 3/2 and 2/3 are the zeroes of the required polynomial.

Let the two zeroes be α and β of the required polynomial.

∴ α = 3/2, β = 2/3

_____________________________

Now,

• Sum of zeroes = α + β

→ (3/2) + (2/3)

→ (9 + 4)/6

→ 13/6

• Product of zeroes = αβ

→ (3/2)(2/3)

→ 1

_____________________________

The required polynomial is :

→ p(x) = k [ x² - (α + β)x + αβ ]

• Putting known values.

→ p(x) = k [ x² - (13/6)x + (1) ]

→ p(x) = k [x² - 13x/6 + 1]

→ p(x) = k [ (6x² - 13x + 6)/6 ]

→ p(x) = k/6 [ 6x² - 13x + 6 ]

• Putting k = 6.

→ p(x) = 6x² - 13x + 6

Answer 2

$\huge{\underline{\underline{\mathfrak\green{Question\::}}}}$

$\sf{zeroes \:of \:a \:polynomial\: are \:3/2\: and\: 2/3 }$ $\sf{then\: find\: the \:polynomial.}$

$\huge{\underline{\underline{\mathfrak\red{Solution\::}}}}$

$\implies$ $\sf{let ,\:the \:two \:zeroes \:be \:α\: and\: β .}$

$\implies$ .°. $\red{ α = 3/2}$ & $\green{β = 2/3}$

$\implies$ $\sf{sum\: of\: zeroes = α + β}$

$\implies$ $\sf{(3/2) + (2/3)}$

$\implies$ $\purple{\boxed{ 13/6}}$

$\implies$ $\sf{product\: of \:zeroes = α × β}$

$\implies$ $\sf{(3/2)×(2/3)}$

$\implies$ $\pink{\boxed{ 1}}$

We know that,

$\implies$ $\bold\orange{p(x) = K[x^2-(α +β)x +α ×β]}$

Now,

$\implies$ $\sf{p(x) = K[x^2\:-(α +β)x\: +α ×β]}$

$\implies$ $\sf{p(x)= K ( x^2-13x/6 +1)}$

$\implies$ $\sf{p(x)= k(6x^2- 13x +6)/6}$

$\implies$ $\sf{p(x) = k/6 ( 6x^2 - 13x +6)}$

$\rightarrow$ $\sf{p(x) = ( 6x^2- 13x +6) \: \: [putting\: K=6. ]}$

$\implies$ $\large\bold\red{\boxed{\boxed{p(x) = 6x^2 -13x +6}}}$