Exercise 2.5 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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EX 2.5
Question 1.
Solve $2 x^2+x-15 \leq 0$.
Solution:
To find the solution of the inequality $a x^2+b x+c \geq 0$ or $a x^2+b x+c \leq 0$ (for $a>0$ )
First we have to solve the quadratic equation $a x^2+\mathrm{bx}+\mathrm{c}=0$
Let the roots be a and $\mathrm{P}$ (where $\mathrm{a}<\mathrm{P}$ )
So for the inequality $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c} \geq 0$ the roots lie outside $\alpha$ and $\beta$ (i.e.,) $x \leq \alpha$ and $x \geq \beta$
So for the inequality $a x^2+b x+c \leq 0$. The roots lie between $\alpha$ and $\beta$ (i.e.,) $x>\alpha$ and $x<\beta$ (i.e.) $a \leq x \leq \beta$
$
\begin{aligned}
& x=\frac{-1 \pm \sqrt{1-4(2)(-15)}}{2(2)}=\frac{-1 \pm \sqrt{121}}{4} \\
& x=\frac{-1 \pm 11}{4} ; x=\frac{-1+11}{4}, \frac{-1-11}{4} \\
& x=\frac{5}{2}, x=-3
\end{aligned}
$
Here $\alpha=-3$ and $\beta=\frac{5}{2}$
(Note that $\alpha<\beta$ )
So for the inequality $2 x^2+x-15 \leq 0$ $x$ lies between -3 and $\frac{5}{2}$
(i.e.) $x \in\left[-3, \frac{5}{2}\right]$ or $-3 \leq x \leq \frac{5}{2}$

Question 2 .
Solve $-x^2+3 x-2 \geq 0$
Solution:
$
-\mathrm{x}^2+3 \mathrm{x}-2 \geq 0 \Rightarrow \mathrm{x}^2-3 \mathrm{x}+2 \leq 0
$
$
(\mathrm{x}-1)(\mathrm{x}-2) \leq 0
$
$[(x-1)(x-2)=0 \Rightarrow x=1$ or 2 . Here $\alpha=1$ and $\beta=2$. Note that $\alpha<\beta]$
So for the inequality $(x-1)(x-2) \leq 2$
$\mathrm{x}$ lies between 1 and 2
(i.e.) $x \geq 1$ and $x \leq 2$ or $x \in[1,2]$ or $1 \leq x \leq 2$