Question

If alpha and beta are the zeroes of the quadratic polynomial f(x)=3x2-4x+1,then find a quadratic polynomial whose zeroes are alpha2/beta and beta2/alpha

Answer 1

Hi Mate !

Here's the Solution :-

• Given equation :- 3x² - 4x + 1

let's factorise it :0 ... by middle term Splitting.

3x² - 4x + 1 = 0

3x² - 3x - x + 1 = 0

3x ( x - 1 ) - 1 ( x - 1 ) = 0

( x - 1 ) ( 3x - 1 ) = 0

° ( x - 1 ) = 0
x = 1

° ( 3x - 1 ) = 0

x = 1/3

$let \: \alpha \: be \: 1 \: and \: \beta \: be \: \frac{1}{3}$

• The new quadratic equation have the zeros as

$\frac{ { \alpha }^{2} }{ \beta } \: and \: \frac{ { \beta }^{2} }{ \alpha }$


=>

$\frac{ ({1}^{2}) }{ \frac{1}{3} } = \frac{1}{ \frac{1}{3} } \\ = 3$
=>

$\frac{ ({ \frac{1}{3}) }^{2} }{1} = \frac{ \frac{1}{9} }{1} \\ = \frac{1}{9}$
So, the Zeros of new equation are 3 and 1/9

♯ Sum of Zeros :-

3 + 1/9

= ( 27 + 1 )/9

= 28/9

♯ Product of Zeros :-

3 × ( 1/9 )

= 1/3

• To form new quadratic equation we have formula as :-

x² - ( Sum of Zeros )x+ ( Product of Zeros)


${x}^{2} - \frac{28}{9} x + \frac{1}{3} = 0$


$\frac{9 {x}^{2} - 28x + 3 }{9} = 0$

9x² - 28x + 3 = 0 ( is the required equation )

Answer 2

For the quadratic equation with roots, α ,β

Sum of roots = 4/3

Product of roots = 1/3 .

Now, The quadratic equation with roots α²/β ,

β²/ α

Now,

Sum of roots = α²/β + β²/ α = α³+β³/αβ

= (α + β)³-3αβ(α+β)/αβ

= (4/3)³-3(1/3)(4/3) / 1/3

= 64/27 - 4/3 / 1/3

= 64/27-36/27 / 1/3

= 28/27 * 3/1

= 28/9

Product of roots = (α²/β * β²/ α) = (αβ ) = 1/3 .

The quadratic equation with required roots = x²-28/9x + 1/3 = 9(x² - 28/8x+1/3) = 9x² - 28x + 3