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Mathematics Question on Quadratic Equations

Let α(a) and β(a)\alpha ( a) \, and \, \beta (a) be the roots of the equation (1+a3−1)x2−(1+a−1)x+(1+a6−1)=0 (\sqrt [3] {1 + a} - 1) x^2 - (\sqrt {1 + a} - 1) x + (\sqrt [ 6] {1 + a} - 1) = 0, where a > - 1. Then, lima→0+lim_{a \to 0^+ }, α(a) and lima→0+β\alpha (a) \, and \, lim_{a \to 0^+ } \beta (a) are

A

−52- \frac{ 5}{2} and 1

B

−12- \frac{ 1}{2} and -1

C

−72- \frac{ 7}{2} and 2

D

−92- \frac{ 9}{2} and 3

Answer

−12- \frac{ 1}{2} and -1

Explanation

Solution

PLAN To make the quadratic into simple form we should eliminate radical sign. Description of Situation As for given equation, when a →\rightarrow 0 the equation reduces to identity in x. i, e., ax2+bx+c=0,∀ x ∈Rax^2 + bx + c = 0, \forall \, x \, \in R or a = b = c →\rightarrow 0 Thus, first we should make above equation independent from coefficients as 0 . Let a+1=t6>a + 1 = t^6> Thus when a→0,t→1a \rightarrow 0, t \rightarrow 1. (t2−1)x2+(t3−1)x+(t−1)=0(t^2 - 1) x^2 + (t^3 - 1) x + (t - 1) = 0 ⇒(t−1)(t+1)x2+(t2+t+1)x+1=0,\Rightarrow (t - 1) \\{ (t + 1) x^2 + ( t^2 + t + 1) x + 1 \\} = 0, as t →1\rightarrow 1 \hspace25mm 2x2+3x+1=02x^2 + 3x + 1 = 0 ⇒\Rightarrow \hspace25mm 2x2+2x+x+1=02x^2 + 2x + x + 1 = 0 ⇒\Rightarrow \hspace25mm ( 2x + 1) (x + 1) = 0 Thus, \hspace25mm x = - 1, - 1 / 2 or \hspace25mm lima→0+α(a)=−1/2lim_{a \to 0^+ } \alpha (a) = -1 / 2 and \hspace25mm lima→0+β(a)=−1lim_{a \to 0^+} \beta (a) = - 1 .