If the roots of the equation a x2-b x-c-0 are real and of the form -1- and -1- then value of a-b-c2 is

The correct option is A b2 4ac Here α+1/α + α 1/α = b/a and α+1/α 1 =c/a Therefore α=(c+a)/(c a) and 2α^2 1/α(α 1) =b/a Substituting α, we get (a+b+c)^2 ...

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Byju's Answer

Standard XII

Mathematics

Nature of Roots of a Cubic Polynomial Using Derivatives

If the roots ...

Question

If the roots of the equation a $x^{2}$ + bx + c = 0 are real and of the form $\frac{(\propto - 1)}{\propto}$ and $\frac{(\propto + 1)}{\propto}$ , then value of ${(a + b + c)}^{2}$ is

b² - 4ac

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b² - 2ac

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2b² - ac

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None

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Solution

The correct option is A
b² - 4ac

Here $\frac{α + 1}{α}$ + $\frac{α - 1}{α}$ = $\frac{- b}{a}$ and $\frac{α + 1}{α - 1}$ = $\frac{c}{a}$

Therefore α= $\frac{(c + a)}{(c - a)}$ and $\frac{2 α^{2} - 1}{α (α - 1)}$ = $\frac{b}{a}$

Substituting α, we get

${(a + b + c)}^{2}$ = $b^{2}$ – 4ac

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If the roots of the equation ax2 + bx + c = 0 are real and of the form(∝−1)∝ and (∝+1)∝, then value of (a+b+c)2 is

The correct option is A b2 - 4ac Here α+1α + α−1α = −ba and α+1α−1 =ca Therefore α=(c+a)(c−a) and 2α2−1α(α−1) =ba Substituting α, we get (a+b+c)2 =b2 – 4ac

If the roots of the equation a $x^{2}$ + bx + c = 0 are real and of the form $\frac{(\propto - 1)}{\propto}$ and $\frac{(\propto + 1)}{\propto}$ , then value of ${(a + b + c)}^{2}$ is