Exercise 2.6 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 2.6 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Ex 2.6
Question 1.
Find the zeros of the polynomial function $f(x)=4 x^2-25$
Solution:
$
\begin{aligned}
4 x^2-25 & =0 \\
4 x^2 & =25 \\
x^2 & =\frac{25}{4} \Rightarrow x= \pm \sqrt{\frac{25}{4}}= \pm \frac{5}{2} \\
x & =-\frac{5}{2}, x=+\frac{5}{2}
\end{aligned}
$
$\therefore$ The zeros of the polynomial are $x= \pm \frac{5}{2}$
Question 2.
If $x=-2$ is one root of $x^3-x^2-17 x=22$, then find the other roots of equation.
Solution:
$x=-2$ is one root
So applying synthetic division

So the other factor is $x^2-3 x-11=0$
$
\begin{aligned}
& x=\frac{3 \pm \sqrt{9+44}}{2(1)} \\
& x=-\frac{3 \pm \sqrt{53}}{2}
\end{aligned}
$
So the roots are $-2, \frac{3 \pm \sqrt{53}}{2}$

Question 3.
Find the real roots of $x^4=16$
Solution:
$
\begin{aligned}
& x^4=16 \\
& \Rightarrow x^4-16=0 \\
& \text { (i.e.,) } x^4-4^2=0 \\
& \Rightarrow\left(x^2+4\right)\left(x^2-4\right)=0
\end{aligned}
$
$x^2+4=0$ will have no real roots
so solving $x^2-4=0$
$
x^2=4
$
$
x= \pm \sqrt{4}= \pm \sqrt{2}
$
So the real roots are $x= \pm 2$
Question 4.
Solve $(2 x+1)^2-(3 x+2)^2=0$
Solution: