Exercise 3.13 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.13$
Question $1 .$
Determine the nature of the roots for the following quadratic equations
(i) $15 \cdot x^{2}+11 \cdot x+2=0$
(ii) $x^{2}-x-1=0$
(iii) $\sqrt{2} t^{2}-3 \mathrm{t}+3 \sqrt{2}=0$
(iv) $9 y^{2}-6 \sqrt{2} y+2=0$
(v) $9 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{x}^{2}-24 \mathrm{abcdx}+16 \mathrm{c}^{2} \mathrm{~d}^{2}=0 \mathrm{a} \neq 0, \mathrm{~b} \neq 0$
Solution:
(i) $15 \mathrm{x}^{2}+11 \mathrm{x}+2=0$ comparing with $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0$.
Here a $=15,6=11, \mathrm{c}=2$
$\begin{aligned}
&\Delta=b^{2}-4 a c \\
&=11^{2}-4 \times 15 \times 2 \\
&=121-120 \\
&=1>1 .
\end{aligned}$
$\therefore$ The roots are real and unequal.
(ii) $\mathrm{x}^{2}-\mathrm{x}-1=0$,
Here $a=1, b=-1, c=-1$.
$\begin{aligned}
&\Delta=b^{2}-4 a c \\
&=(-1)^{2}-4 \times 1 \times-1 \\
&=1+4=5>0 .
\end{aligned}$
$\therefore$ The roots are real and unequal.
(iii) $\sqrt{2} t^{2}-3 \mathrm{t}+3 \sqrt{2}=0$
Here $\mathrm{a}=\sqrt{2}, \mathrm{~b}=-3, \mathrm{c}=3 \sqrt{2}$
$\begin{aligned}
&\Delta=b^{2}-4 a c \\
&=(-3)^{2}-4 \times \sqrt{2} \times 3 \sqrt{2}
\end{aligned}$

$=9-24=-15<0 \text {. }$
$\therefore$ The roots are not real.
(iv) $9 \mathrm{y}^{2}-6 \sqrt{2} y+2=0$
$\mathrm{a}=9, \mathrm{~b}=6 \sqrt{2}, \mathrm{c}=2$
$\Delta=\mathrm{b}^{2}-4 \mathrm{ac}$
$=(6 \sqrt{2})^{2}-4 \times 9 \times 2$
$=36 \times 2-72$
$=72-72=0$
(v) $9 a^{2} b^{2} x^{2}-24 a b c d x+16 c^{2} d^{2}=0$
$\Delta=\mathrm{b}^{2}-4 \mathrm{ac}$
$=(-24 a b c d)^{2}-4 \times 9 a^{2} b^{2} \times 16 c^{2} d^{2}$
$=576 a^{2} b^{2} c^{2} d^{2}-576 a^{2} b^{2} c^{2} d^{2}$
$=0$

Question $2 .$
Find the value(s) of 'A' for which the roots of the following equations are real and equal.
(i) $(5 \mathrm{k}-6) \mathrm{x}^{2}+2 \mathrm{kx}+\mathrm{l}=0$
Answer:
Here $\mathrm{a}=5 \mathrm{k}-6 ; \mathrm{b}=2 \mathrm{k}$ and $\mathrm{c}=1$
Since the equation has real and equal roots $\Delta=0$.
$\therefore \mathrm{b}^{2}-4 \mathrm{ac}=0$
$\begin{aligned}
&(2 \mathrm{k})^{2}-4(5 \mathrm{k}-6)(1)-0 \\
&4 \mathrm{k}^{2}-20 \mathrm{k}+24=0
\end{aligned}$
$(\div 4) \Rightarrow \mathrm{k}^{2}-5 \mathrm{k}+6=0$
$(\mathrm{k}-3)(\mathrm{k}-2)=0$
$k=3=0$ or $k-2=0$

$k=3$ or $k=2$
The value of $k=3$ or 2
(ii) $k x^{2}+(6 \mathrm{k}+2) \mathrm{x}+16=0$
Answer:
Here $\mathrm{a}=\mathrm{k}, \mathrm{b}=6 \mathrm{k}+2 ; \mathrm{c}=16$
Since the equation has real and equal roots
$\begin{aligned}
&\Delta=0 \\
&\mathrm{~b}^{2}-4 \mathrm{ac}=0 \\
&(6 \mathrm{k}+2)^{2}-4(\mathrm{k})(16)=0 \\
&36 \mathrm{k}^{2}+4+24 \mathrm{k}-4(\mathrm{k})(16)=0 \\
&36 \mathrm{k}^{2}-40 \mathrm{k}+4=0 \\
&(\div \mathrm{by} 4) \Rightarrow 9 \mathrm{k}^{2}-10 \mathrm{k}+1=0 \\
&9 \mathrm{k}^{2}-9 \mathrm{k}-\mathrm{k}+1=0 \\
&9 \mathrm{k}(\mathrm{k}-1)-1(\mathrm{k}-1)=0 \\
&9 \mathrm{k}(\mathrm{k}-1)-1(\mathrm{k}-1)=0 \\
&(\mathrm{k}-1)(9 \mathrm{k}-1)=0 \\
&\mathrm{k}-1 \text { or } 9 \mathrm{k}-1=0 \\
&\mathrm{k}=1 \text { or } \mathrm{k}=\frac{1}{9}
\end{aligned}$
The value of $k=1$ or $\frac{1}{9}$

Question $3 .$
If the roots of $(a-b) x^{2}+(b-c) x+(c-a)=0$ are real and equal, then prove that $b, a, c$ are in arithmetic progression.
Solution:
$\begin{aligned}
&(\mathrm{a}-\mathrm{b}) \mathrm{x}^{2} \text { । }(\mathrm{b}-\mathrm{c}) \mathrm{x} \mid(\mathrm{c}-\mathrm{a})-0 \\
&\mathrm{~A}=(\mathrm{a}-\mathrm{b}), \mathrm{B}=(\mathrm{b}-\mathrm{c}), \mathrm{C}=(\mathrm{c}-\mathrm{a}) \\
&\Delta=\mathrm{b}^{2}-4 \mathrm{ac}=0 \\
&\Rightarrow(\mathrm{b}-\mathrm{c})^{2}-4(\mathrm{a}-\mathrm{b})(\mathrm{c}-\mathrm{a}) \\
&\Rightarrow \mathrm{b}^{2}-2 \mathrm{bc}+\mathrm{c}^{2}-4\left(\mathrm{ac}-\mathrm{bc}-\mathrm{a}^{2}+\mathrm{ab}\right) \\
&\Rightarrow \mathrm{b}^{2}-2 \mathrm{bc}+\mathrm{c}^{2}-4 \mathrm{ac}+4 \mathrm{bc}+4 \mathrm{a}^{2}-4 \mathrm{ab}=0 \\
&\Rightarrow 4 \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+2 \mathrm{bc}-4 \mathrm{ac}-4 \mathrm{ab}=0 \\
&\left.\Rightarrow-(-2 \mathrm{a}+\mathrm{b}+\mathrm{c})^{2}=0\left[\because(\mathrm{a}+\mathrm{b}+\mathrm{c})=\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+2 \mathrm{ab}+2 \mathrm{bc}+2 \mathrm{ca}\right)\right] \\
&\Rightarrow 2 \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\
&\Rightarrow 2 \mathrm{a}=\mathrm{b}+\mathrm{c} \\
&\therefore \mathrm{a}, \mathrm{b}, \mathrm{c} \text { are in A.P. }
\end{aligned}$

Question $4 .$
If $\mathrm{a}, \mathrm{b}$ are real then show that the roots of the equation
$(a-b) x^{2}-6(a+b) x-9(a-b)=0$ are real and unequal.
Answer:
$\begin{aligned}
&(a-b) x^{2}-6(a+b) x-9(a-b)=0 \\
&\text { Here } a=a-b ; b=-6(a+b) ; c=-9(a-b) \\
&\Delta=b^{2}-4 a c
\end{aligned}$

$\begin{aligned}
&=[-6(a+b)]^{2}-4(a-b)[-9(a-b)] \\
&=36(a+b)^{2}+36(a-b)(a-b) \\
&=36(a+b)^{2}+36(a-b)^{2} \\
&=36\left[(a+b)^{2}+(a-b)^{2}\right]
\end{aligned}$
The value is always greater than 0
$\Delta=36\left[(a+b)^{2}+(a-b)^{2}\right]>0$
$\therefore$ The roots are real and unequal.
 

Question $5 .$
If the roots of the equation $\left(c^{2}-a b\right) x^{2}-2\left(a^{2}-b c\right) x+b^{2}-a c=0$ are real and equal prove that either $a=0$ (or) $a^{3}+b^{3}+c^{3}=3 a b c$.
Solution:
$\left(c^{2}-a b\right) x^{2}-2\left(a^{2}-b c\right) x+\left(b^{2}-a c\right)-0$
$\Delta=\mathrm{B}^{2}-4 \mathrm{AC}=0$ (since the roots are real and equal)
$\begin{aligned}
&\Rightarrow 4\left(a^{2}-b c\right)^{2}-4\left(c^{2}-a b\right)\left(b^{2}-a c\right)=0 \\
&\Rightarrow 4\left(a^{4}-2 a^{2} b c+b^{2} c^{2}\right)-4\left(c^{2} b^{2}-a b^{3}-a c^{3}+a^{2} b c\right)=0 \\
&\Rightarrow 4 a^{4}+4 b^{2} c^{2}-8 a^{2} b c-4 c^{2} b^{2}+4 a b^{3}+4 a c^{3}-4 a^{2} b c=0 \\
&\Rightarrow 4 a^{4}+4 a b^{3}+4 a c^{3}-4 a^{2} b c-8 a^{2} b c=0
\end{aligned}$
$\begin{aligned}
&\Rightarrow 4 a\left[a^{3}+b^{3}+c^{3}\right]=0 \text { or } a=0 \\
&\Rightarrow a=0 \text { or }\left[a^{3}+b^{3}+c^{3}-3 a b c\right]=0 \\
&\Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=0
\end{aligned}$
$\Rightarrow \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=3 \mathrm{abc}$ or $\mathrm{a}=0$
Hence proved.