Solveeit Logo
Login

Question

Mathematics Question on Complex Numbers and Quadratic Equations

Let α,β be the roots of the equation, ax2+bx+c=0.a,b,c are real and sn=αn+βn and ∣31+s11+s2\1+s11+s21+s3\1+s21+s31+s4∣=k(a+b+c)2a4\begin{vmatrix}3 &1+s_1 &1+s_2\\\1+s_1&1+s_2 &1+s_3\\\1+s_2&1+s_3 &1+s_4\end{vmatrix}=\frac{k(a+b+c)^2}{a^4} then k=

A

b2-4ac

B

b2+4ac

C

b2+2ac

D

4ac-b2

Answer

b2-4ac

Explanation

Solution

The correct answer is option (A): b2-4ac

∣31+s11+s2\1+s11+s21+s3\1+s21+s31+s4∣\begin{vmatrix}3 &1+s_1 &1+s_2\\\1+s_1&1+s_2 &1+s_3\\\1+s_2&1+s_3 &1+s_4\end{vmatrix}=∣1+1+11+α+β1+α2+β2 1+α+β1+α2+β21+α3+β3 1+α2+β21+α3+β31+α4+β4∣\begin{vmatrix} 1+1+1 &1+\alpha+\beta &1+\alpha^2+\beta^2\\\ 1+\alpha+\beta&1+\alpha^2+\beta^2 &1+\alpha^3+\beta^3 \\\ 1+\alpha^2+\beta^2&1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4 \end{vmatrix}

=∣111 1αβ 1α2β2∣∣111 1αβ 1α2β2∣=\begin{vmatrix} 1 &1 &1 \\\ 1&\alpha & \beta \\\ 1&\alpha^2 &\beta^2 \end{vmatrix}\begin{vmatrix} 1 &1 &1 \\\ 1&\alpha & \beta \\\ 1&\alpha^2 &\beta^2 \end{vmatrix}

=[αβ(β−α)−β2+α2+β−α][αβ(β−α)−(β2−α2)+(β−α)]=[\alpha\beta(\beta-\alpha)-\beta^2+\alpha^2+\beta-\alpha][\alpha\beta(\beta-\alpha)-(\beta^2-\alpha^2)+(\beta-\alpha)]

=(β−α)2[αβ(β+α)+1]2=(\beta-\alpha)^2[\alpha\beta(\beta+\alpha)+1]^2………..(i)

Now, ax2+bx+c=0

here α+β=−ba\alpha+\beta=-\frac{b}{a} and αβ=ca\alpha\beta=\frac{c}{a}

Now, (α−β)2=(α+β)2−4αβ(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta

\Rightarrow$$\frac{b^2}{a^2}-\frac{4c}{a} = b2−4aca2\frac{b^2-4ac}{a^2}

Now, Δ\Delta = (b2−4aca2)[ca−(−ba)+1](\frac{b^2-4ac}{a^2})[\frac{c}{a}-(-\frac{b}{a})+1]

⇒(b2−4aca2)(c+a+ba)2\Rightarrow (\frac{b^2-4ac}{a^2})(\frac{c+a+b}{a})^2 = 1a4(b2−4ac)(a+b+c)2\frac{1}{a^4}(b^2-4ac)(a+b+c)^2

Now, k(a+b+c)2a4=1a4(b2−4ac)(a+b+c)2\frac{k(a+b+c)^2}{a^4} =\frac{1}{a^4}(b^2-4ac)(a+b+c)^2

⇒k=(b2−4ac)\Rightarrow k = (b^2-4ac)