If$$f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma $$, where $$\alpha $$,$$\beta $$,$$\gamma $$ are rational numbers and two roots of $$f(x) = 0$$ and eccentricities of a parabola and a rectangular hyperbola, then $$\alpha + \beta + \gamma $$ is equal to:A.-1B.0C.1D.2

Question

If$$f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma $$, where $$\alpha $$,$$\beta $$,$$\gamma $$ are rational numbers  and two roots of $$f(x) = 0$$ and eccentricities of a parabola and a rectangular hyperbola, then $$\alpha  + \beta  + \gamma $$ is equal to:A.-1B.0C.1D.2

Vedantu Content Team · Accepted Answer

Hint: By the given equation and condition we find the roots of the given polynomial. After finding the roots of the polynomial we solve this and comparing with the given polynomial, we obtain the $$\alpha $$,$$\beta $$,$$\gamma $$ values.Complete step-by-step answer:We know that the eccentricity of a parabola is $$1$$.Also, the eccentricity of a rectangular hyperbola can be determined and it’s shown below:Rectangular hyperbola is a hyperbola for which the asymptotes are perpendicular, also called an equilateral hyperbola or right hyperbola. This occurs when the si-major and semi- major are equal.So $$a = b$$We know that the eccentricity of hyperbola$$e = \dfrac{{{{({a^2} + {b^2})}^{\dfrac{1}{2}}}}}{a}$$In rectangular hyperbola we have $$a = b$$$$ \Rightarrow e = \dfrac{{{{(2{a^2})}^{\dfrac{1}{2}}}}}{a}$$$$ \Rightarrow e = \dfrac{{\sqrt 2 a}}{a}$$Hence, $$e = \sqrt 2 $$Thus the root of polynomial $$f(x)$$ is $$1$$, $$\sqrt 2 $$ and $$ - \sqrt 2 $$Now $$f(x) = (x - 1)(x - \sqrt 2 )(x + \sqrt 2 )$$Expanding the first and second brackets,$$ \Rightarrow f(x) = ({x^2} - x\sqrt 2  - x + \sqrt 2 )(x + \sqrt 2 )$$Again expanding barkers, we get a polynomial,$$ \Rightarrow f(x) = {x^3} - {x^2}\sqrt 2  - {x^2} + \sqrt 2 x + {x^2}\sqrt 2  - 2x - \sqrt 2 x + 2$$Cancelling terms and rearranging,$$ \Rightarrow f(x) = {x^3} - {x^2} - 2x + 2$$Comparing with the coefficients given polynomial, $$f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma $$We get that $$\alpha  =  - 1$$, $$\beta  =  - 2$$ and $$\gamma  = 2$$Now we need the value of $$\alpha  + \beta  + \gamma $$, and we know the individual values,Substituting these we get,$$ \Rightarrow \alpha  + \beta  + \gamma  =  - 1 - 2 + 1$$$$ \Rightarrow \alpha  + \beta  + \gamma  =  - 1$$So, the correct answer is “Option A”.Note: In conic section, there is a locus of a point in which the distance to the point and the line are in the constant ratio. That ratio is known as eccentricity. It is denoted by$$e$$. We choose another root as $$ - \sqrt 2 $$ because if one root is irrational it occurs in a conjugate pair.

If\[f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma \], where \[\alpha \],\[\beta \],\[\gamma \] are rational numbers and two roots of \[f(x) = 0\] and eccentricities of a parabola and a rectangular hyperbola, then \[\alpha + \beta + \gamma \] is equal to:A.-1B.0C.1D.2

If\[f(x) = {x^3} + \alpha {x^2} + \beta x + \gamma \], where \[\alpha \],\[\beta \],\[\gamma \] are rational numbers and two roots of \[f(x) = 0\] and eccentricities of a parabola and a rectangular hyperbola, then \[\alpha + \beta + \gamma \] is equal to:
A.-1
B.0
C.1
D.2