Question

if alpha and beeta are the zeroes of (ax^2+bx+c)evaluate alpha minus beeta​

Answer 1

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Answer 2

Answer :-

α - β = √(b² - 4ac) / a

Solution :-

ax² + bx + c

Let α, β be the zeroes of the polynomial

Sum of zeroes = α + β = - b/a

Product of zeroes = αβ = - b/a

By using identity (α - β)² = (α + β)² - 4αβ

$\implies ( \alpha - \beta )^{2} = ( \alpha + \beta )^{2} - 4 \alpha \beta$

Substituting the values

$\implies ( \alpha - \beta )^{2} = \bigg( - \dfrac{b}{a} \bigg)^{2} - 4 \bigg( \dfrac{c}{a} \bigg)$

$\implies ( \alpha - \beta )^{2} = \dfrac{( - b)^{2} }{a^{2} } - \dfrac{4c}{a}$

$\implies ( \alpha - \beta )^{2} = \dfrac{b^{2} }{a^{2} } - \dfrac{4c}{a}$

Taking LCM

$\implies ( \alpha - \beta )^{2} = \dfrac{b^{2} - 4ac}{a^{2} }$

Taking Square root on both sides

$\implies \sqrt{( \alpha - \beta )^{2}} = \sqrt{ \dfrac{b^{2} - 4ac}{a^{2} } }$

$\implies \alpha - \beta = \dfrac{ \sqrt{ b^{2} - 4ac}}{ \sqrt{ {a}^{2} } }$

$\implies \alpha - \beta = \dfrac{ \sqrt{ b^{2} - 4ac}}{a }$

Therefore the value of α - β is √(b² - 4ac) / a