Exercise 3.5-Additional Problems - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Solve: $3^{2 \mathrm{x}+4}+1=2.3^{\mathrm{x}+2}$
Solution:
$
\begin{aligned}
& 3^{2 x+4}=3^{x+2}+3^{x+2}-1 \\
& 3^{2 x+4}-3^{x+2}=3^{x+2}-1 \Rightarrow 3^{x+2}\left[3^{x+2}-1\right]=\left[3^{x+2}-1\right] \\
& 3^{x+2}=1 \Rightarrow 3^{x+2}=3^0 \\
& x+2=0 \Rightarrow x=-2
\end{aligned}
$
Question 2.
Solve: $2^x-2^{x+3}+2^4=0$.
Solution:
$
\begin{aligned}
& 2^{2 x}-\left(2^x \cdot 2^3\right)+2^4=0 \text { since } 2^3=8=4+4=2^2+2^2 \\
& 2^{2 x}-(4+4) 2^x+2^4=0 \Rightarrow 2^{2 x}-\left(2^2+2^2\right) 2^x+2^4=0 \\
& \left(2^x-2^2\right)\left(2^x-2^2\right)=0 \Rightarrow\left(2^x-2^2\right)^2=0 \\
& 2^x=2^2 \Rightarrow x=2
\end{aligned}
$
Question 3.
Solve: $(x-4)(x+2)(x+3)(x-3)+8=0$.
Solution:
$
\begin{aligned}
& (x-4)(x+3)(x+2)(x-3)+8=0 \\
& \left(x^2-x-12\right)\left(x^2-x-6\right)+8=0
\end{aligned}
$
Let $y=x^2-x$
$(\mathrm{y}-12)(\mathrm{y}-6)+8=0 \Rightarrow \mathrm{y}^2-18 \mathrm{y}+72+8=0$
$
\mathrm{y}^2-18 \mathrm{y}+80=0 \Rightarrow(\mathrm{y}-10)(\mathrm{y}-8)=0
$
Case (i) $y-10=0$
$
x^2-x-8=0
$

Case (ii)
$
\begin{aligned}
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{1 \pm \sqrt{1+40}}{2}=\frac{1 \pm \sqrt{41}}{2} \\
y-8 & =0 \Rightarrow x^2-x-8=0 \\
x & =\frac{1 \pm \sqrt{1+32}}{2}=\frac{1 \pm \sqrt{33}}{2}
\end{aligned}
$
Question 4.
Solve: $(x+y)^{2 / 3}+2(x-y)^{2 / 3}=3\left(x^2-y^2\right)^{1 / 3}$ and $3 x-2 y=13$.
Solution:

$
\begin{gathered}
(x+y)^{2 / 3}+2(x-y)^{2 / 3}=3\left(x^2-y^2\right)^{1 / 3} \\
3 x-2 y=13
\end{gathered}
$
$\div\left(x^2-y^2\right)^{1 / 3} \Rightarrow\left(\right.$ i.e) dividing by $(x-y)^{1 / 3}(x+y)^{1 / 3}$
$
\begin{aligned}
\frac{(x+y)^{\frac{2}{3}}}{(x-y)^{\frac{1}{3}}(x+y)^{\frac{1}{3}}}+\frac{2(x-y)^{\frac{2}{3}}}{(x-y)^{\frac{1}{3}}(x+y)^{\frac{1}{3}}}=3 & \\
\frac{(x+y)^{\frac{1}{3}}}{(x-y)^{\frac{1}{3}}}+2 \frac{(x-y)^{\frac{1}{3}}}{(x+y)^{\frac{1}{3}}}=3 & \\
\text { Let } z=\left(\frac{x+y}{x-y}\right)^{\frac{1}{3}} & \\
z^2+2=3 z & \Rightarrow z+\frac{2}{z}=3 \\
(z-2)(z-1)=0 & \Rightarrow z^2+3 z+2=0 \\
& \Rightarrow z=2, z=1
\end{aligned}
$

Question 5.
$
5^{x-1}+5^{1-x}=26
$
Solution:
$
\begin{aligned}
& \frac{5^x}{5}+\frac{5}{5^x}=26 \\
& \text { Let } y=5^x \\
& \frac{y}{5}+\frac{5}{y}=26 \quad \Rightarrow y^2+25=26 y \\
& y^2-26 y+25=0 \quad \Rightarrow(y-1)(y-25)=0 \\
& y-1=0, y-25=0 \\
&
\end{aligned}
$

Question 6.
Solve: $12 x^4-56 x^3+89 x^2-56 x+12=0$
Solution:
Since the co-efficients of the equations are equal from both ends.
Divide the equation by $x^2$

$\begin{aligned}
12 x^2-56 x+89-\frac{56}{x}+\frac{12}{x^2} & =0 \\
12\left(x^2+\frac{1}{x^2}\right)-56\left(x+\frac{1}{x}\right)+89 & =0 \\
x+\frac{1}{x} & =y \\
12\left(y^2-2\right)-56 y+89 & =0 \\
12 y^2-24-56 y+89 & =0 \\
12 y^2-56 y+65 & =0 \\
(2 y-5)(6 y-13) & =0
\end{aligned} \quad \Rightarrow x^2+\frac{1}{x^2}=y^2-2$

Question 7.
Solve: $(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=\mathbf{1 0}$
Solution:

$
(\sqrt{3}+\sqrt{2})^x(\sqrt{3}-\sqrt{2})^x=10
$
We have $(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1$
$
\begin{aligned}
& (\sqrt{3}-\sqrt{2})=\frac{1}{\sqrt{3}+\sqrt{2}} \\
& \therefore \quad(1) \Rightarrow(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10 \text {. } \\
& \text { Let } y=(\sqrt{3}+\sqrt{2})^x \\
& y+\frac{1}{y}=10 \\
& y^2-10 y+1=0 \\
& y=\frac{10 \pm \sqrt{100-4}}{2} \\
& y=\frac{10 \pm \sqrt{96}}{2}=\frac{10 \pm 2 \sqrt{24}}{2}=5 \pm \sqrt{24} \\
& y=5+\sqrt{24}=(\sqrt{3}+\sqrt{2})^2 \\
& y=\pm(\sqrt{3}+\sqrt{2}) \\
& (\sqrt{3}+\sqrt{2})^x=(\sqrt{3}+\sqrt{2})^2 \Rightarrow x=2 \\
& (\sqrt{3}-\sqrt{2})^x=(\sqrt{3}-\sqrt{2})^{-2} \Rightarrow x=-2 \\
& \therefore \quad x=-2,2 \\
&
\end{aligned}
$

Question 8.
Solve: $x^4+4 x^3+5 x^2+4 x+1=0$
Solution:
$
\begin{aligned}
x^2+4 x+5+\frac{4}{x}+\frac{1}{x^2} & =0 \\
\left(x^2+\frac{1}{x^2}\right)+4\left(x+\frac{1}{x}\right)+5 & =0 \\
\operatorname{Let} y & =\left(x+\frac{1}{x}\right) \Rightarrow x^2+\frac{1}{x^2}=y^2-2 \\
\left(y^2-2\right)+4 y+5 & =0 \\
y^2+4 y+3 & =0 \\
(y+3)(y+1) & =0
\end{aligned}
$