Exercise 3.11 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.11$
Question $1 .$
Solve the following quadratic equations by completing the square method
(i) $9 x^{2}-12 x+4=0$
(ii) $\frac{5 x+7}{x-1}=3 x+2$
Solution:
(i)
$\begin{array}{ll}
9 x^{2}-12 x+4=0 \div 9 & \left(\frac{-b}{2 a}\right)^{2}=\left(\frac{-2}{3}\right)^{2} \\
\frac{9 x^{2}}{9}-\frac{12}{9} x+\frac{4}{9}=0 & {\left[\frac{4}{2(3)}\right]^{2}=\left(\frac{2}{3}\right)^{2}} \\
x^{2}-\frac{4}{3} x=\frac{-4}{9} &
\end{array}$
$\begin{aligned}
&a=1, b=-\frac{4}{3} \\
&x^{2}-\frac{4}{3} x+\left(\frac{+2}{3}\right)^{2}=\frac{-4}{9}+\left(\frac{+2}{3}\right)^{2}
\end{aligned}$
[Adding $\left(\frac{2}{3}\right)^{2}$ both sides]

$$
\begin{aligned}
\left(x-\frac{2}{3}\right)^{2} &=\frac{-4}{9}+\frac{4}{9}=0 \\
x-\frac{2}{3} &=0, x-\frac{2}{3}=0 \\
x &=\frac{2}{3}, \frac{2}{3}
\end{aligned}
$$
(ii)
$\frac{5 x+7}{x-1}=3 x+2$
$5 x+7=(x-1)(3 x+2)$
$5 x+7=3 x^{2}-3 x+2 x-2$
$3 x^{2}-x-5 x-2-7=0$
$\begin{aligned} 3 x^{2}-6 x-9 &=0 \div 3 \\ x^{2}-2 x-3 &=0 \\ x^{2}-2 x &=3 \end{aligned} \quad \begin{aligned}&a=1, b=-2 \\&\left(\frac{-b}{2 a}\right)^{2}=\left(\frac{2}{2}\right)^{2}=1^{2}\end{aligned}$
$x^{2}-2 x+1=3+1$
[Adding 1 both sides]
$\begin{aligned}
(x-1)^{2} &=4 \\
(x-1) &=\sqrt{4}=\pm 2 \\
x-1=2 \Rightarrow x &=3 \\
x-1=-2 \Rightarrow x &=-1
\end{aligned}$

Question 2.
Solve the following quadratic equations by formula method
(i) $2 x^{2}-5 x+2=0$
(ii) $\sqrt{2} f^{2}-6 f+3 \sqrt{2}$
(iii) $3 y^{2}-20 y-3=0$
(iv) $36 y^{2}-12 a y+\left(a^{2}-b^{2}\right)=0$
Solution:

(i) $2 x^{2}-5 x+2=0$
The formula for finding roots of a quadratic equation $a x^{2}+b x+c=0$ is
$\begin{aligned}
x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
\frac{2 x^{2}-5 x+2}{a} c &=0 \\
\therefore x &=\frac{-(-5) \pm \sqrt{(-5)^{2}-4 \times 2 \times 2}}{2 \times 2} \\
&=\frac{5 \pm \sqrt{25-16}}{4} \\
&=\frac{5 \pm \sqrt{9}}{4}=\frac{5 \pm 3}{4}=\frac{8}{4}, \frac{2}{4} \\
&=2, \frac{1}{2} \\
1 &
\end{aligned}$
$\therefore$ Solutions is $2, \frac{1}{2}$

(ii) $\underset{a}{\sqrt{2}} f^{2}-\underset{b}{b}+\underset{c}{3}+\sqrt{2}=0$
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Here $f=\frac{-(-6) \pm \sqrt{(-6)^{2}-4 \times \sqrt{2} \times 3 \sqrt{2}}}{2 \times \sqrt{2}}$
$=\frac{6 \pm \sqrt{36-24}}{2 \sqrt{2}}$
$=\frac{6 \pm \sqrt{12}}{2 \sqrt{2}}=\frac{6 \pm 2 \sqrt{3}}{2 \sqrt{2}}=\frac{\not 2(3 \pm \sqrt{3})}{22 \sqrt{2}}$
$\Rightarrow \frac{3+\sqrt{3}}{\sqrt{2}}, \frac{3-\sqrt{3}}{2}$
(iii) $3 y^{2}+20 y-23=0$
$a \quad b \quad c$
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Here $y=\frac{-(-20) \pm \sqrt{(-20)^{2}-4 \times 3 \times-23}}{2 \times 3}$
$=\frac{20 \pm \sqrt{400+276}}{6}$

$\begin{aligned}
&=\frac{20 \pm \sqrt{676}}{6}=\frac{20 \pm 26}{6} \\
&=\frac{46}{6} \text { or } \frac{-6}{6} \\
y & \Rightarrow \frac{23}{3} \text { or }-1
\end{aligned}$
(iv) 36
$36 y^{2}-12 a y+\left(a^{2}-b^{2}\right)=0$
$a$
$b$
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Here $y=\frac{-(-12 a) \pm \sqrt{(-12 a)^{2}-4 \times 36 \times\left(a^{2}-b^{2}\right)}}{2 \times 36}$
$=\frac{12 a \pm \sqrt{144 a^{2}-144 a^{2}+144 b^{2}}}{72}$
$=\frac{12 a \pm 12 b}{72} \Rightarrow \frac{a \pm b}{6}$
$=\frac{a+b}{6}, \frac{a-b}{6}$
 

Ouestion $3 .$
A ball rolls down a slope and travels a distance $\mathrm{d}=\mathrm{t}^{2}-0.75 \mathrm{t}$ feet in $\mathrm{t}$ seconds. Find the time when the distance travelled by the ball is $11.25$ feet.
Solution:
Distance $\mathrm{d}=\mathrm{t}^{2}-0.75 \mathrm{t}$,
Given that $\mathrm{d}-11.25-\mathrm{t}^{2}-0.75 \mathrm{t}$.

$\begin{aligned}
t^{2}-0.75 \mathrm{t}-11.25=0 \\
t &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&=\frac{(+0.75) \pm \sqrt{(-0.75)^{2}-4 \times 1 \times-11.25}}{2 \times 1} \\
&=\frac{+0.75 \pm \sqrt{0.5625+45}}{2} \\
&=\frac{+0.75 \pm \sqrt{45.5625}}{2} \\
&=\frac{+0.75 \pm 6.75}{2} \\
&=\frac{7.50}{2} \text { or } \frac{-6}{2} \\
&=3.75 \text { or }-3 \text { It is not possible. } \\
\therefore t &=3.75 \mathrm{~s} .
\end{aligned}$