Exercise 2.4 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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EX 2.4
Question 1.
Construct a quadratic equation with roots 7 and -3 .
Solution:
The given roots are 7 and -3
Let $\alpha=7$ and $\beta=-3$
$\alpha+\beta=7-3=4$
$\alpha \beta=(7)(-3)=-21$
The quadratic equation with roots $\alpha$ and $\beta$ is $\mathrm{x}^2-(\alpha+\beta) \mathrm{x}+\alpha \beta=0$
So the required quadratic equation is
$\mathrm{x}^2-(4) \mathrm{x}+(-21)=0$
(i.e.) $x^2-4 x-21=0$
Question 2.
A quadratic polynomial has one of its zeros $1+\sqrt{5}$ and it satisfies $\mathrm{p}(1)=2$. Find the quadratic polynomial.
Solution:
Given $\alpha=1+\sqrt{5}$ So, $\beta=1-\sqrt{5}$
$\alpha+\beta=2 ; \alpha \beta=1^2-(-\sqrt{5})^2=1-5=-4$
The quadratic polynomial is
$
\begin{aligned}
& \mathrm{p}(\mathrm{x})=\mathrm{x}^2-(\alpha+\beta) \mathrm{x}+\alpha \beta \\
& \mathrm{p}(\mathrm{x})=\mathrm{k}\left(\mathrm{x}^2-2 \mathrm{x}-4\right) \\
& \mathrm{p}(1)=\mathrm{k}(1-2-4)=-5 \mathrm{k} \\
& \text { Given } \mathrm{p}(1)=2 \\
& \Rightarrow-5 k=2 \\
& \Rightarrow \quad k=\frac{-2}{5} \\
& \therefore p(x)=\frac{-2}{5}\left(x^2-2 x-4\right)
\end{aligned}
$
Question 3.
If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+\sqrt{2} \mathrm{x}+3=0$, form a quadratic polynomial with zeroes $1 / \alpha, 1 / \beta$.
Solution:
$\alpha$ and $\beta$ are the roots of the equation $x^2+\sqrt{2} x+3=0$

$
\begin{gathered}
\Rightarrow \alpha+\beta=-\frac{\sqrt{2}}{1}=-\sqrt{2} \text { and } \alpha \beta=\frac{3}{1}=3 \\
\text { (i.e.) } \alpha+\beta=-\sqrt{2} \text { and } \alpha \beta=3
\end{gathered}
$
The quadratic equation with roots
$\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is $x^2-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) x+\left(\frac{1}{\alpha} \times \frac{1}{\beta}\right)=0$
Now $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-\sqrt{2}}{3}$
$
\frac{1}{\alpha} \times \frac{1}{\beta}=\frac{1}{\alpha \beta}=\frac{1}{3}
$
So the required quadratic equation is
$
\begin{aligned}
& x^2-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) x+\left(\frac{1}{\alpha} \times \frac{1}{\beta}\right)=0 \\
& \text { (i.e, } x^2-\left(\frac{-\sqrt{2}}{3}\right) x+\frac{1}{3}=0 \\
& x^2+\frac{\sqrt{2}}{3} x+\frac{1}{3}=0 \Rightarrow(\div \text { by } 3) \\
& 3 x^2+\sqrt{2} x+1=0
\end{aligned}
$
Question 4.
If one root of $\mathrm{k}(\mathrm{x}-1)^2=5 \mathrm{x}-7$ is double the other root, show that $\mathrm{k}=2$ or -25 .
Solution:
$
\begin{aligned}
& \mathrm{k}(\mathrm{x}-1)^2=5 \mathrm{x}-7 \\
& \left(\text { i.e..) } \mathrm{k}\left(\mathrm{x}^2-2 \mathrm{x}+1\right)-5 \mathrm{x}+7=0\right. \\
& \mathrm{x}^2(\mathrm{k})+\mathrm{x}(-2 \mathrm{k}-5)+\mathrm{k}+1=0 \\
& \mathrm{kx}^2-\mathrm{x}(2 \mathrm{k}+5)+(\mathrm{k}+7)=0
\end{aligned}
$
Here it is given that one root is double the other.
So let the roots to $\alpha$ and $2 \alpha$

Sum of the roots $=\alpha+2 \alpha=3 \alpha=\frac{2 k+5}{k} \Rightarrow \alpha=\frac{2 k+5}{3 k} \ldots$
Product of the roots $=\alpha(2 \alpha)=2 \alpha^2=\frac{k+7}{k}$
$
\Rightarrow \alpha^2=\frac{k+7}{2 k}
$
Substituting $\alpha$ value from (1) in (2)
we get $\left(\frac{2 k+5}{3 k}\right)^2=\frac{k+7}{2 k}$
$
\begin{aligned}
& \frac{4 k^2+25+20 k}{9 k^2}=\frac{k+7}{2 k} \\
& 2\left(4 \mathrm{k}^2+25+20 \mathrm{k}\right)=9 \mathrm{k}(\mathrm{k}+7) \\
& 2\left(4 \mathrm{k}^2+25+20 \mathrm{k}\right)=9 \mathrm{k}^2+63 \mathrm{k} \\
& 8 \mathrm{k}^2+50+40 \mathrm{k}-9 \mathrm{k}^2-63 \mathrm{k}=0 \\
& -\mathrm{k}^2-23 \mathrm{k}+50=0 \\
& \mathrm{k}^2+23 \mathrm{k}-5 \mathrm{o}=0 \\
& (\mathrm{k}+25)(\mathrm{k}-2)=0 \\
& \mathrm{k}=-25 \text { or } 2
\end{aligned}
$
Question 5 .
If the difference of the roots of the equation $2 x^2-(a+1) x+a-1=0$ is equal to their product then prove that $\mathrm{a}=2$.
Solution:
$
2 x^2-(a+1) x+(a-1)=0
$
Let $\alpha$ and $\beta$ be the roots $\Rightarrow \alpha+\beta=\frac{a+1}{2}$ and $\alpha \beta=\frac{a-1}{2}$
$
\begin{aligned}
\Rightarrow \quad & (\alpha+\beta)^2=(\alpha \beta)^2 \\
& (\text { i.e., })(\alpha+\beta)^2-4 \alpha \beta=(\alpha \beta)^2 \\
\Rightarrow \quad & \left(\frac{a+1}{2}\right)^2-4\left(\frac{a-1}{2}\right)=\left(\frac{a-1}{2}\right)^2 \\
& \frac{a^2+2 a+1}{4}-2(a-1)=\frac{a^2-2 a+1}{4} \\
& a^2+2 a+1-8(a-1)=a^2-2 a+1 \\
& a^2+2 a+1-8 a+8-a^2+2 a-1=0
\end{aligned}
$

$
\begin{aligned}
-4 a+8=0 & \Rightarrow 4 a=8 \\
a & =\frac{8}{4}=2
\end{aligned}
$
Question 6.
Find the condition that one of the roots of $a x^2+b x+c$ may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be $\alpha$ and $-\beta$
Sum of the roots $=-b / a=0 \Rightarrow b=0$
(ii) Let the roots be $\alpha, 3 \alpha$
Sum of the roots $=4 \alpha=-b / a \Rightarrow \alpha=\frac{-b}{4 a}$
Product of the roots $=3 \alpha^2=\frac{c}{a} \Rightarrow 3\left(\frac{-b}{4 a}\right)^2=c / a$ $3\left(\frac{b^2}{16 a^2}\right)=c / a \Rightarrow 3 b^2 \times a=16 a^2 c$.
$(\div$ by $a) 3 b^2=16 a c$
(iii) Let the roots be $\alpha, \frac{1}{\alpha}$
Product of the roots $=\alpha \times \frac{1}{\alpha}=\frac{c}{a} \Rightarrow 1=\frac{c}{a} \Rightarrow c=a$

Question 7.
If the equations $\mathrm{x}^2-\mathrm{ax}+\mathrm{b}=0$ and $\mathrm{x}^2-\mathrm{ex}+\mathrm{f}=0$ have one root in common and if the second equation has equal roots that $a e=2(b+f)$.
Solution:

Question 8.

Discuss the nature of roots of
(i) $-x^2+3 x+1=0$
(ii) $4 x^2-\mathrm{x}-2=0$
(iii) $9 x^2+5 x=0$
Solution:
(i) $-x^2+3 x+1=0$
$\Rightarrow$ comparing with $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$
$\Delta=\mathrm{b}^2-4 \mathrm{ac}=(3)^2-4(1)(-1)=9+4=13>0$
$\Rightarrow$ The roots are real and distinct
(ii) $4 \mathrm{x}^2-\mathrm{x}-2=0$
$\mathrm{a}=4, \mathrm{~b}=-1, \mathrm{c}=-2$
$\Delta=\mathrm{b}^2-4 \mathrm{ac}=(-1)^2-4(4)(-2)=1+32=33>0$
$\Rightarrow$ The roots are real and distinct
(iii) $9 \mathrm{x}^2+5 \mathrm{x}=0$
$\mathrm{a}=9, \mathrm{~b}=5, \mathrm{c}=0$
$\Delta=\mathrm{b}^2-4 \mathrm{ac}=5^2-4(9)(0)=25>0$
$\Rightarrow$ The roots are real and distinct
Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the $\mathrm{x}-$ axis and if so in how many points.
(i) $y=x^2+x+2$
(ii) $y=x^2-3 x-1$
(iii) $y=x^2+6 x+9$
Solution:

(i) $y=x^2+x+2$
$
\begin{aligned}
y & =\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^{2+2}=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}+2 \\
& =\left(x+\frac{1}{2}\right)^2+\frac{7}{4}
\end{aligned}
$
Then $\mathrm{D}=b^2-4 a c=(1)^2-4(1)(2)=1-8=-7$
$\therefore \mathrm{D}$ is - ve. The parabola will never intersect $x$ axis.
The minimum point is $\left(-\frac{1}{2}, \frac{7}{4}\right)$
(ii)
$
\begin{aligned}
y & =x^2-3 x-7=\left(x-\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-7 \\
& =\left(x-\frac{3}{2}\right)^2-\frac{9}{4}-7=\left(x-\frac{3}{2}\right)^2-\frac{37}{4}
\end{aligned}
$
Then $\mathrm{D}=(-3)^2-4(1)(-7)=9+28=37$
$\therefore \mathrm{D}$ is $+\mathrm{ve}$. The parabola intersect $x$ axis at two point.
The minimum point is $\left(\frac{3}{2},-\frac{37}{4}\right)$
(iii)
$
\begin{aligned}
y & =x^2+6 x+9 \\
& =(x+3)^2
\end{aligned}
$
Then $D=(6)^2-4(1)(9)=36-36=0$
$\therefore \mathrm{D}$ is zero. The parabola intersect $x$ axis at $(-3,4)$.
The minimum point is $(-3,0)$
Question 10.
Write $f(x)=x^2+5 x+4$ in completed square form.

Solution:
$
\begin{aligned}
x^2+5 x+4 & =x^2+2\left(\frac{5}{2}\right) x+4+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2 \\
& =x^2+2\left(\frac{5}{2}\right) x+\frac{25}{4}+4-\frac{25}{4} \\
& =\left(x+\frac{5}{2}\right)^2-\frac{9}{4}=\left(x+\frac{5}{2}\right)^2-\left(\frac{3}{2}\right)^2 \\
y & =\left(x+\frac{5}{2}\right)^2-\left(\frac{3}{2}\right)^2
\end{aligned}
$