Question

if aphla and beat are the zeros of the quadratic polynomial f(x)=x2-1 find quadratic polynomial whose zeros are 2aphla /beat and 2beat /aphla​

Answer 1

The required polynomial is $\bf x^2+4x+4$

Accurate Question :

If α and β are zeroes of polynomial $\bf f(x)=x^2-1$ ,Find a quadratic polynomial whose zeroes are $\bf \dfrac{2 \alpha}{ \beta}\;and\;\dfrac{2 \beta}{ \alpha}$

Solution :

Given :

α and β are zeroes of the polynomial $\bf f(x)=x^2-1$

• First Compare $\bf f(x)= x^2-1 \;with\;\;ax^2+bx+c$

So,

• a = 1
• b = 0
• c = - 1

$\rule{300}{1.5}$

1)

$\implies \sf \alpha+\beta\\ \\ \\\implies \sf \dfrac{-b}{a}\\ \\ \\ \implies \sf \dfrac{0}{1}\\ \\ \\ \implies \sf 0\qquad[Eq\;1]$

2)

$\implies \sf \alpha\beta\\ \\ \\ \implies \sf \dfrac{c}{a}\\ \\ \\ \implies \sf \dfrac{-1}{1}\\ \\ \\ \implies \sf -1\qquad[Eq\;2]$

3)

$\implies \sf \dfrac{2\alpha}{\beta}+\frac{2\beta}{\alpha}\\ \\ \\\implies \sf \dfrac{2(\alpha^{2}+\beta^{2})}{\alpha\beta}\\ \\ \\\implies \sf \dfrac{2[(\alpha+\beta)^{2}-2\alpha\beta]}{\alpha\beta}\\ \\ \\\implies \sf \dfrac{2[0-2(-1)]}{(-1)}\\ \\ \\\implies \sf \dfrac{4}{(-1)}\\ \\ \\ \implies \sf -4\qquad[Eq\;3]$

4)

$\implies \sf \left(\dfrac{2\alpha}{\beta}\right)\left(\dfrac{2\beta}{\alpha}\right)\\ \\ \\\implies \sf 4\qquad[Eq\;4]$

$\rule{300}{1.5}$

∴ The quadratic polynomial :

$\implies \sf k\bigg[x^{2}-\bigg(\dfrac{2\alpha}{\beta}+\dfrac{2\beta}{\alpha}\bigg)+\bigg(\dfrac{2\alpha}{\beta}\bigg)\bigg(\dfrac{2\beta}{\alpha}\bigg)\bigg] \\ \\ \\\textbf{\bigg[Where k is a constant.\bigg] }\\ \\ \\ \implies \sf k[x^{2}-(-4)x+4\\ \\ \\\implies \sf k(x^{2}+4x+4)$

So,

We can put any value of k,

So, k = 1

Therefore, The required polynomial is $\bf x^2+4x+4$