# Quadratic Inequalities Study Material For IIT JEE

Quadratic Inequalities or Inequations is an important concept included in Quadratic Equation IIT JEE Mathematics syllabus. Direct questions are generally asked from this concept in both JEE Mains and JEE Advanced. We, at Byju’s, understand the importance and necessity of having proper revision notes. Therefore, we have come up with an all-inclusive IIT JEE Quadratic Equation Revision notes designed by our expert faculty members to help students in boosting their IIT JEE preparations.

Wavy Curve Method or Methods of Intervals:

The Wavy Curve Method or the Methods of Intervals is helpful in solving Inequalities of the form:

$\mathbf{\frac{F\;(x)}{G\;(x)}\;>\; 0,\;\frac{F\;(x)}{G\;(x)}\geq \; 0,\;\frac{F\;(x)}{G\;(x)}\;<\; 0,\frac{F\;(x)}{G\;(x)}\leq \; 0 }$

Step 1: Consider the given Polynomial Equation and Find all the roots of the given polynomial Equation F (x) and G (x).

i.e. $\mathbf{H\;(x)\;=\;\frac{F\;(x)}{G\;(x)}\;=\; \frac{(x\;-\alpha _{1})\;(x\;-\alpha _{2})\;(x\;-\alpha _{3})\;.\;.\;.\;.\;.\;.\;(x\;-\alpha _{n})}{(x\;-\eta _{1})\; (x\;-\eta _{2})\;(x\;-\eta _{3})\;.\;.\;.\;.\;.\;(x\;-\eta _{m})}}$

Where, $α_{1}, α_{2}, α_{3}, . . . . . . . . α_{n}$ are roots of F (x) and $β_{1}, β_{2}, β_{3}, . . . . . . . . β_{m}$ are roots of polynomial equation G (x).

Note: H (x) = 0 for x = $α_{1}, α_{2}, α_{3}, . . . . . . . . α_{n}$ and H (x) is not defined for x = $β_{1}, β_{2}, β_{3}, . . . . . . . . β_{n}$

Step 2: Compare the roots of both F (x) and G (x) and arrange all the roots of F (x) and G (x) in increasing order say $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, . . . . . . . . . a_{m}$+ n.

Step 3: Plot them on the number line. Now, draw wavy curve starting from the right of am + n along the number line that alternatively changes its position at these points.

Note: H (x) is a positive function for all the intervals in which the curve lies above the number line and H (x) is a negative function for all the intervals in which the curve lies below the number line.

Note: All the zeros of the given polynomial equation H (x) must be marked with colored black circles on the number line whereas, all points of discontinuities of the function H (x) must be marked on the number line with white circles.

If ax2 + bx + c > 0 and ($a \neq 0$):

Case 1: (i) If D (b2 – 4ac) > 0 i.e. the quadratic equation f (x) has two different roots and a < b

Then, if a > 0 then, $\mathbf{x \in \;(-\infty , \alpha )\cup \; \left ( \eta ,\infty \right )}$

And, if a < 0 then, x $\mathbf{ \in }$ (α, β)

Case 2: If D (b2 – 4ac) = 0, i.e. the quadratic equation f (x) has equal roots i.e. a = b.

Then, if a > 0 then, $\mathbf{x \in \;(-\infty , \alpha )\cup\; \left ( \alpha ,\infty \right )}$

And, if a < 0 then $\mathbf{x \in }$ Ø

Case 3: If D (b2 – 4ac) < 0, i.e. the quadratic equation has imaginary roots

Then, if a > 0 then, $\mathbf{x \in \;R}$

And, if a < 0 then, $\mathbf{x \in}$Ø

In General, if (x – a) (x – b) ≥ 0, then a ≤ x ≤ b,

(x – a) (x – b) ≤ 0 and a < b then a ≤ x or x ≥ b.

Hence, the quadratic Inequalities can be quickly solved using the method of intervals.

Example 1: Let P (x) = $\mathbf{\frac{(x\;-\;3)\;(x\;+\;2)}{(x\;-\;7)\;(x\;+\;1)}}$. Find the values of P (x) for which the given function is positive or negative.

Solution:

On arranging the roots of the given polynomial equation in increasing order [-2, -1, +3, +7] and plotting them on the number line.

P (x) is positive i.e. P (x) > 0, then $\mathbf{x \in \;(-\;1,\;3)}$

P (x) is negative i.e. P (x) < 0, then $\mathbf{x \in \;(-\;2,\;-\;1)\;(3,\; 7)}$

Example 2: Solve the given Quadratic In equation:

$\mathbf{f\;(x)\;=\;\frac{3\;x^{2}\;+\;8\;-\;7x}{x^{2}\;+\;1}\leq\; 2}$

Solution:

Given, $\mathbf{\frac{3\;x^{2}\;+\;8\;-\;7x}{x^{2}\;+\;1}\leq\; 2}$

i.e. $\mathbf{f\;(x)\;=\;\frac{3\;x^{2}\;+\;8\;-\;7x}{x^{2}\;+\;1}\;-2\leq\; 0}$

Or, $\mathbf{\frac{3\;x^{2}\;+\;8\;-\;7x\;-\;2\;-2\;x^{2}}{x^{2}\;+\;1}\leq\; 0}$

Or, $\mathbf{\frac{x^{2}\;+\;6\;-\;7x}{x^{2}\;+\;1}\leq\; 0}$

Or, $\mathbf{\frac{\left ( x\;-\;6 \right )\;\left ( x\;-\;1 \right )}{x^{2}\;+\;1}\leq\; 0}$

Neglecting $x^{2} + 1$ as x $\mathbf{\in }$ R

Therefore, x $\mathbf{\in }$ [1, 6]

Example 3: Solve the given Quadratic In equation:

$\mathbf{-\;\frac{1}{x\;-\;2}\geq \;\frac{1}{x}\;+\;\frac{2}{x\;+\;2} }$

Solution:

From the given quadratic Inequality x can’t be 2, – 2, 0.

The given quadratic inequality can be rewritten as:

$\mathbf{\frac{1}{x\;-\;2}\;-\;\frac{1}{x}\;-\;\frac{2}{x\;+\;2}\leq \;0 }$

i.e. $\mathbf{\frac{x\;(x\;+\;2)\;-\;(x\;+\;2)\;(x\;-\;2)\;-\;2x\;(x\;-\;2)}{x\;(x\;-\;2)\;(x\;+\;2)}\leq \;0}$

Or, $\mathbf{\frac{x^{2}\;+\;2x\;-\;x^{2}\;+\;4\;-\;2x^{2}\;+\;4x}{x\;(x\;-\;2)\;(x\;+\;2)}\leq \;0}$

Or, $\mathbf{\frac{-\;2x^{2}\;+\;6x\;+\;4}{x\;(x\;-\;2)\;(x\;+\;2)}\leq \;0}$

Or, $\mathbf{\frac{x^{2}\;-\;3x\;-\;2}{x\;(x\;-\;2)\;(x\;+\;2)}\geq \;0}$

i.e. $\mathbf{\frac{\left [x\;- \;\left ( \frac{+\;3\;-\;\sqrt{9\;+\;8}}{2} \right ) \right ]\;\left [x\;-\; \left ( \frac{+\;3\;+\;\sqrt{9\;+\;8}}{2} \right ) \right ]}{x\;(x\;-\;2)\;(x\;+\;2)}\geq \;0}$

Or, $\mathbf{\frac{\left [x\;- \;\left ( \frac{3\;-\;\sqrt{17}}{2} \right ) \right ]\;\left [x\;-\; \left ( \frac{3\;+\;\sqrt{17}}{2} \right ) \right ]}{x\;(x\;-\;2)\;(x\;+\;2)}\geq \;0}$

On arranging the roots of the given polynomial equation in increasing order and plotting them on the number line.

$\mathbf{-\;2 \; \rightarrow \; \frac{3\;-\;\sqrt{17}}{2}\; \rightarrow \; 0\; \rightarrow \;2\; \rightarrow \;\frac{3\;+\;\sqrt{17}}{2}}$

Therefore, from the above graph:

$\mathbf{x\in \;\left ( -\;2,\; \frac{3\;-\;\sqrt{17}}{2} \right ]\cup \;(0,\;2)\cup \;\left [ \frac{3\;+\;\sqrt{17}}{2},\;+\infty \right )}$

Example 4: Find the values of ‘m’ satisfying the given quadratic inequality $\mathbf{\left [ x\in\; R \right ]}$:

$\mathbf{\left | \;\frac{x^{2}\;+\;mx\;+\;1}{x^{2}\;+\;x\;+1}\; \right |\;<\; 2}$

Solution:

Given, $\mathbf{\left | \;\frac{x^{2}\;+\;mx\;+\;1}{x^{2}\;+\;x\;+1}\; \right |\;<\; 2}$

Since, if |x| < k, then – k < x < + k

i.e. $\mathbf{-\;2\;< \;\left | \;\frac{x^{2}\;+\;mx\;+\;1}{x^{2}\;+\;x\;+1}\; \right |\;<\; 2}$ . . . . . . . (1)

The given Quadratic Equation x2 + x + 1 can be further rewritten as $\mathbf{\left ( x\;+\;\frac{1}{2} \right )^{2}\;+\;\frac{3}{4}}$

Therefore, $x^{2} + x + 1$ is positive for all the values of x.

On simplifying Equation (1) we will get:

– 2($x^{2} + x + 2$) < $x^{2} + kx + 1$ < 2 ($x^{2} + x + 1$),

i.e. – 2($x^{2} + x + 2$ ) < $x^{2} + kx + 1$ and $x^{2} + kx + 1$ < 2 ($x^{2} + x + 1$),

Or, 3×2 + x (2 + k) + 3 > 0 and $x^{2}$ + x (2 – k) + 1 > 0

Discriminant of both the Equations must be less than zero for both the quadratic equations to be positive for all the values of x.

i.e. (2 + k)2 – 36 < 0 and (2 – k)2 – 4 < 0,

Or, (k – 4) (k + 8) < 0 and (k – 4).k < 0,

i.e. – 8 < k < 4 and 0 < k < 4.

Hence, k $\mathbf{\in }$ (0, 4)

Example 5: Solve the given Quadratic Inequality:

$\mathbf{\left | \;\frac{x^{2}\;-\;3x\;-\;1}{x^{2}\;+\;x\;+\;1} \; \right | \;< \;3}$

Solution:

Since $x^{2} + x + 1$ will always be greater than zero for all values of x.

Therefore, f (x) = $\mathbf{\frac{\left |\;x^{2}\;-\;3x\;-\;1 \; \right |}{x^{2}\;+\;x\;+\;1}\;<\;3\; }$

i.e. | $x^{2} – 3x – 1$ | < 3 ($x^{2} + x + 1$)

On squaring both the sides,

i.e. $(x^{2} – 3x – 1)^{2} – [ 3(x^{2} + x + 1) ]^{2} < 0$,

Or, $(x^{2} – 3x – 1 + 3\times 2 + 3x + 3) (x^{2} – 3x – 1 – 3\times 2 – 3x -3) < 0$

Or, $(4 \times 2 + 2) (- 2 \times 2 – 6x – 4) < 0,$

Or, (2 \times 2 + 1) (x^{2} + 3x + 2) > 0,

Or, (2 \times 2 + 1) (x + 1) (x + 2) > 0,

Therefore, $\mathbf{x\in \;(-\infty,\; -\;2)\;(-\;1, \infty ) }$