Exercise 3.4-Additional Problems - Chapter 3 - Theory of Equations - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Questions 1.
Solve $(x-3)(x-6)(x-1)(x+2)+54=0$.
Solution:
$
\begin{aligned}
& (x-3)(x-1)(x-6)(x+2)+54=0 \\
& \left(x^2-4 x+3\right)\left(x^2-4 x-12\right)+54=0 \\
& \text { Put } x^2-4 x=y \\
& (y+3)(y-12)+54=0 \\
& y^2-9 y-36+54=0 \\
& y^2-9 y+18=0 \\
& (y-3)(y-6)=0
\end{aligned}
$

Question 2.
Solve the equation $(x-4)(x-2)(x-1)(x+1)+8=0$
Solution:
The equation can be rewritten as $\{(x-4)(x+1)\}\{(x-2)(x-1)\}+8=0$ $\left(\mathrm{x}^2-3 \mathrm{x}-4\right)\left(\mathrm{x}^2-3 \mathrm{x}+2\right)+8=0$
Let $y=x^2-3 x$
$(y-4)(y+2)+8=0 \quad \Rightarrow y^2-2 y-8+8=0$
Case $(i)$
$x=0, x=3$
Case (ii)
$x^2-3 x=2 \quad \Rightarrow x^2-3 x-2=0$
$x=\frac{3 \pm \sqrt{17}}{2}$