Question

if alfa and beta are the zeroes of the x^2-2x -8 then form a quadratic equation whose zeroes are 2alfa and 2 beta

Answer 1

Find the attached file for your query !

Answer 2

Answer :

The given polynomial is (x² - 2x - 8) whose zeroes are α and β

Then,

α² - 2α - 8 = 0 ...(i)

β² - 2β - 8 = 0 ...(ii)

In order to find the required quadratic equation whose roots are 2α and 2β, let us consider that

y = 2α

⇨ α = $\frac{y}{2}$

Now, substituting this value in (i), we get

${(\frac{y}{2})}^{2}-2(\frac{y}{2})-8=0$

$\implies \frac{{y}^{2}}{4}-y-8=0$

$\implies \frac{{y}^{2}-4y-32}{4}=0$

⇨ y² - 4y - 32 = 0, which is the required quadratic equation whose roots are 2α and 2β

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