Exercise 3.4 - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.4$
Question 1.
Solve:
(i) $(x-5)(x-7)(x+6)(x+4)=504$
(ii) $(x-4)(x-7)(x-2)(x+1)=16$
Solution:
(i) $(x-5)(x+4)(x-7)(x+6)=504$
$\left(x^2-x-20\right)\left(x^2-x-42\right)=504$
Let $y=\left(x^2-x\right)$
$(\mathrm{y}-20)(\mathrm{y}-42)=504$
$\Rightarrow \mathrm{y}^2-42 \mathrm{y}-20 \mathrm{y}+840=504$
$\Rightarrow \mathrm{y}^2-62 \mathrm{y}+336=0$
$\Rightarrow(\mathrm{y}-56)(\mathrm{y}-6)=0$
$\Rightarrow(\mathrm{y}-56)=0$ or $(\mathrm{y}-6)=0$
$\Rightarrow \mathrm{x}^2-\mathrm{x}-56=0$ or $\mathrm{x}^2-\mathrm{x}-6=0$
$\Rightarrow(\mathrm{x}-8)(\mathrm{x}+7)=0$ or $(\mathrm{x}-3)(\mathrm{x}+2)=0$
$\Rightarrow \mathrm{x}=8,-7$ or $\mathrm{x}=3,-2$
The roots are $8,-7,3,-2$
(ii) $(\mathrm{x}-4)(\mathrm{x}-7)(\mathrm{x}-2)(\mathrm{x}+1)=16$
$\Rightarrow(\mathrm{x}-4)(\mathrm{x}-2)(\mathrm{x}-7)(\mathrm{x}+1)=16$
$\Rightarrow\left(x^2-6 x+8\right)\left(x^2-6 x-7\right)=16$
Let $x^2-6 x=y$
$(\mathrm{y}+8)(\mathrm{y}-7)=16$
$\Rightarrow \mathrm{y}^2-7 \mathrm{y}+8 \mathrm{y}-56-16=0$
$\Rightarrow \mathrm{y}^2+\mathrm{y}-72=0$
$\Rightarrow(\mathrm{y}+9)(\mathrm{y}-8)=0$
$y+9=0$
$x^2-6 x+9=0$
$(\mathrm{x}-3)^2=0$
$\mathrm{x}=3,3$
or
$
\begin{aligned}
& y-8=0 \\
& x^2-6 x-8=0
\end{aligned}
$

$
\begin{aligned}
& x=\frac{6 \pm \sqrt{36-4(1)(-8)}}{2} \\
& x=\frac{6 \pm \sqrt{36+32}}{2} \\
& x=\frac{6 \pm \sqrt{68}}{2}=\frac{6 \pm 2 \sqrt{17}}{2} \\
& x=\frac{2(3 \pm \sqrt{17})}{2}=3 \pm \sqrt{17}
\end{aligned}
$
The roots are $3,3,3 \pm \sqrt{17}$

Question 2.
Solve: $(2 x-1)(x+3)(x-2)(2 x+3)+20=0$.
Solution:
$
\begin{aligned}
& (2 x-1)(2 x+3)(x+3)(x-2)+20=0 \\
& \Rightarrow\left(4 x^2+6 x-2 x-3\right)\left(x^2-2 x+3 x-6\right)+20=0 \\
& \Rightarrow\left(4 x^2+4 x-3\right)\left(x^2+x-6\right)+20=0 \\
& \Rightarrow\left[4\left(x^2+x\right)-3\right]\left[x^2+x-6\right]+20=0
\end{aligned}
$
Let $y=x^2+x$
$
\begin{aligned}
& \Rightarrow(4 y-3)(y-6)+20=0 \\
& \Rightarrow 4 y^2-24 y-3 y+18+20=0 \\
& \Rightarrow 4 y^2-27 y+38=0 \\
& \Rightarrow(4 y-19)(y-2)=0 \\
& (4 y-19)=0 \\
& 4\left(x^2+x\right)-19=0 \\
& 4 x^2+4 x-19=0 \\
& x=\frac{-4 \pm \sqrt{16+4(4) 19}}{8} \\
& x=\frac{-4 \pm \sqrt{16+304}}{8} \\
& x=\frac{-4 \pm \sqrt{320}}{8}=\frac{-4 \pm 8 \sqrt{5}}{8} \\
& \text { or } \\
& (y-2)=0 \\
& x^2+x-2=0 \\
& (x+2)(x-1)=0
\end{aligned}
$
$
\mathrm{x}=-2,+1
$
The roots are $-2,1, \frac{-1 \pm 2 \sqrt{5}}{2}$