Exercise 2.6 - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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: