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Question

If $$\alpha $$ and $$\beta $$ are zeroes of the quadratic polynomial $$4{x^2} + 4x + 1$$, then find quadratic polynomial whose zeroes are $$2\alpha $$ and $$2\beta $$.


Solution

to solve this lets divide the quadratic equation by 4 such that we have it in the form $$x^{2}+(\alpha+\beta)x+\alpha\beta$$

so after dividing by 4 we have
$$\dfrac{4x^{2}+4x+1}{4}$$
$$x^{2}+x+\frac{1}{4}$$

So we have$$ \alpha+\beta=1$$               ...................(1)
and$$ \alpha\beta=\dfrac{1}{4}$$                                ..................(2)

So if zeroes to the new quadratic equation are $$2\alpha$$ and $$ 2\beta$$, then 

$$2\alpha+2\beta = 2(\alpha+\beta)$$
$$= 2(1)=2$$                        .................... Using (1)

and,
$$2\alpha\times 2\beta= 4\alpha\beta $$
$$=4\times \dfrac{1}{4}= 1$$                      .................... Using (2)

So the new quadratic equation with its roots $$2\alpha\, and \, 2\beta$$ will be
$$x^{2}+(2\alpha+2\beta)x+2\alpha\times2\beta$$

Putting the values, the equation would be,
$$x^{2} + 2x+1$$

Mathematics

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