CAT 2025 Lesson : Inequalities - Quadratic Inequalities

What values of $x$ satisfy $3x^2 + 8 \gt 10x$?

(1) (-$\infty$, 4/3)             
(2) (2, $\infty$)              
(3) (4/3, 2)            
(4) Both (1) and (2)

Solution

$3x^2 -10x +8 \gt 0$

⇒ $3x^2 -6x -4x +8 \gt 0$

⇒ $3x(x-2) -4(x-2) \gt 0$

⇒ $(3x-4) (x-2) \gt 0$

Deflection points are $3x - 4 = 0$ ⇒ $\bm{x = \dfrac{4}{3}}$ and $x - 2 = 0$ ⇒ $\bm{x = 2}$

Marking the deflection points on a number line, the region in the right-most will be positive. The sign for the preceding regions will alternate between positive and negative.



For $(3x - 4)(x - 2) \gt 0,$ the LHS should be positive
∴ Acceptable values of $x$ are when $x \lt \dfrac{4}{3}$ or when $x \gt 2$

Answer:($4$) Both (1) and (2)

What are the acceptable values for $x$ where $\dfrac {x+5}{x-1} \lt 2$?

Solution

To solve for an inequality, we need to have 0 on the RHS.

$\dfrac{x+5}{x-1} \lt 2$ ⇒ $\dfrac{x+5}{x-1} -$2$\lt 0$ ⇒ $\dfrac{-x+7}{x-1} \lt 0$

As the coefficient of $x$ should be +ve in all factors, we shall multiply LHS and RHS by –1.

$\dfrac{x-7}{x-1} \gt 0$------(1)

It does not matter if the linear terms (or factorized terms) are multiplied or divided, the treatment is the same in an inequality. We find the deflection points for each term. Deflection points are at $x = 1, 7$.



The LHS in (1) needs to be positive. This happens when $x < 1$ or $x > 7$.

Answer: $x < 1$ or $x > 7$

What values of $x$ satisfy $2x^3-x^2-7x+6 \lt 0$?
($1$) $(-∞ ,1)$       
($2$) $(-2,1)$       
($3$) $(-2,1) ∪ (\dfrac{3}{2},∞)$       
($4$) $(-∞ ,-2) ∪ (1,\dfrac{3}{2})$      

Solution

Let $f(x)$ = $2x^3-x^2-7x+6$

Substituting small integral values in $x$ we get $f(1)$ = $2-1-7+6$ = $0$

Dividing by ($x-1$), we get ($2x^2+x-6$).

∴ $f(x) = 2x^3-x^2-7x+6 = (x-1) (2x^2+x-6)$

= $(x-1) (2x-3) (x+2)$

$2x^3-x^2-7x+6 \lt 0$

⇒ $(x-1) (2x-3) (x+2) \lt 0$



The negative regions $(-∞ ,-2)$ and $(1,\dfrac{3}{2})$

Answer: ($4$) $(-∞ ,-2) ∪ (1,\dfrac{3}{2})$

What is the range of $x$ values that satisfy $\dfrac {x^2 - 7x + 22}{x + 2} \gt 2$?

Solution

$\dfrac {x^2 - 7x + 22}{x + 2} \gt 2$ ⇒ $\dfrac {x^2 - 7x + 22}{x + 2} - 2\gt 0$

⇒ $\dfrac {x^2-9x+18}{x+2} \gt 0$

⇒ $\dfrac {(x-6)(x-3)}{(x+2)} \gt 0$



Inequality is positive when $x$ is in the range $(-2,3) ∪ (6,∞)$.

Answer: ($4$) $(-2,3) ∪ (6,∞)$