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

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Additional Questions
Question 1.
Find the values of $\mathrm{k}$ so that the equation $\mathrm{x}^2=2 \mathrm{x}(1+3 \mathrm{k})+7(3+2 \mathrm{k})=0$ has real and equal roots. Solution:
The equation is $\mathrm{x}^2-\mathrm{x}(2)(1+3 \mathrm{k})+7(3+2 \&)=0$
The roots are real and equal $\Rightarrow \Delta=0$ (i.e.,) $b^2-4 \mathrm{ac}=0$
Here $\mathrm{a}=1, \mathrm{~b}=-2(1+3 \mathrm{k}), \mathrm{c}=7(3+2 \mathrm{k})$
So $\mathrm{b}^2-4 \mathrm{ac}=0 \Rightarrow[-2(1+3 \mathrm{k})]^2-4(1)(7)(3+2 \mathrm{k})=0$
(i.e.,) $4(1+3 \mathrm{k})^2-28(3+2 \mathrm{k})=0$
$(\div$ by 4$)(1+3 \mathrm{k})^2-7(3+2 \mathrm{k})=0$
$1+9 \mathrm{k}^2+6 \mathrm{k}-21-14 \mathrm{k}=0$
$9 \mathrm{k}^2-8 \mathrm{k}-20=0$
$(\mathrm{k}-2)(9 \mathrm{k}+10)=0$
$\Rightarrow \quad k-2=0$ or $9 k+10=0$
$\Rightarrow \quad k=2$ or $k=\frac{-10}{9}$
To solve the quadratic inequalities $a x^2+b x+c<0$ (or) $a x^2+b x+c>0$
Question 2.
If the sum and product of the roots of the quadratic equation $\operatorname{ax}^2-5 x+c=0$ are both equal to 10 then find the values of $\mathrm{a}$ and $\mathrm{c}$.
Solution:
The given equation is $\mathrm{ax}^2-5 \mathrm{x}+\mathrm{c}=0$

$\text { Let the roots be } \alpha \text { and } \beta \text { Given } \alpha+\beta=10 \text { and } \alpha \beta=10$

Question 3.
If $\alpha$ and $\beta$ are the roots of the equation $3 x^2-4 x+1=0$, form the equation whose roots are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$
Solution:
$\alpha$ and $\beta$ are the roots of $3 x^2-4 x+1=0$
$
\Rightarrow \alpha+\beta=-\left(\frac{-4}{3}\right)=\frac{4}{3} \text { and } \alpha \beta=\frac{1}{3}
$
The new roots are $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$
$
\begin{aligned}
\text { Sum of the roots } & =\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta} \\
& =\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}=\frac{\left(\frac{4}{3}\right)^3-3\left(\frac{1}{3}\right)\left(\frac{4}{3}\right)}{\frac{1}{3}} \\
& =\frac{\frac{64}{27}-\frac{4}{3}}{1 / 3}=3\left(\frac{64-36}{27}\right) \\
& =\frac{3(28)}{27}=\frac{28}{9}
\end{aligned}
$
Product of the roots $=\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha}=\alpha \beta=\frac{1}{3}$
So the quadratic equation is with roots $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$
$
\begin{aligned}
x^2-\left(\frac{28}{9}\right) x+\left(\frac{1}{3}\right) & =0 \\
x^2-\frac{28 x}{9}+\frac{1}{3} & =0 \\
\Rightarrow \quad 9 x^2-28 x+3=0 &
\end{aligned}
$

Question 4.
If one root of the equation $3 \mathrm{x}^2+\mathrm{kx}-81=0$ is the square of the other then find $\mathrm{k}$.
Solution:
Let the roots be $\alpha$ and $\alpha^2$
Sum of the roots $\alpha+\alpha^2=\frac{-k}{3} \quad \ldots$ (1)
Product of the roots $\alpha\left(\alpha^2\right)=\left(\alpha^3\right)=\frac{-81}{3}=-27$ $\alpha^3=-27=(-3)^3 \Rightarrow \alpha=-3$
substituting $\alpha$ value in (1) we get
$
-3+(-3)^2=\frac{-k}{3}
$
(i.e.)
$
\frac{-k}{3}=6 \Rightarrow k \Rightarrow \begin{gathered}
3 \\
-18
\end{gathered}
$
Question 5.
If one root of the equation $2 x^2-a x+64=0$ is twice that of the other then find the value of a

Solution:
Let the roots be $\alpha, 2 \alpha$
Sum of the roots $\alpha+2 \alpha=3 \alpha=\frac{-(-a)}{2}=\frac{a}{2}$
$
\Rightarrow \alpha=\frac{a}{6}
$
Product of the roots $=\alpha(2 \alpha)=2 \alpha^2=\frac{64}{2}=32$
$
\Rightarrow \alpha^2=\frac{32}{2}=16 \Rightarrow \alpha=\sqrt{16}= \pm 4
$
Substituting $\alpha$ value in (1) we get
$
\pm 4=\frac{a}{6} \Rightarrow a= \pm 6 \times 4= \pm 24
$