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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