$\alpha ,\beta $ are roots of the equation $2{x^2} - 5x - 6 = 0$ then
- the equation whose roots are $\alpha ,\beta $ and 2 is $2{x^3} - 9{x^2} + 4x + 12 = 0$
- The value of $\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha } + 1} \right)$ is $\frac{{ - 37}}{{12}}$
- the equation whose roots are $\alpha ,\beta $ and -1 is $2{x^3} - 3{x^2} - 11x - 6 = 0$
- The value of ${(\alpha - \beta )^2}$ is $\frac{{53}}{4}$
Reveal answer
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C
Correct answer
Explanation
Roots of 2x^2 - 5x - 6 = 0 are alpha and beta. Sum = 5/2, Product = -3. If roots are alpha, beta, and -1, the cubic equation is (x + 1)(2x^2 - 5x - 6) = 2x^3 - 5x^2 - 6x + 2x^2 - 5x - 6 = 2x^3 - 3x^2 - 11x - 6 = 0.
AI explanation
If the roots of the new cubic equation are alpha, beta, and -1, then the required equation can be found by multiplying the original quadratic by the factor x plus 1. Multiplying 2x squared minus 5x minus 6 by x plus 1 gives 2x cubed minus 3x squared minus 11x minus 6. The result is the equation whose roots are alpha, beta and -1 is 2x^3 - 3x^2 - 11x - 6 = 0.