Quadratic Inequalities

Solving Quadratic Inequalities

Wavy Curve Method or Methods of Intervals:

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

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

Steps For Solving Quadratic Inequalities

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.

Quadratic Inequalities

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 coloured 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.

Shortcut Tips For Solving Quadratic Inequalities

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.

Solution of the Inequality

(a) Write all the terms present in the inequality as their linear factors in standard form i.e. x + a.

(b) If the inequality contains quadratic expression, f(x) = ax2 + bx + c; then first check the discriminant (D = b2 – 4ac)

(i) If D > 0, then the expression can be written as f(x) = a (x – α)(x – β). Where α and β are given by

\(\alpha ,\beta =\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\)

(ii) If D = 0, then the expression can be written as \(f(x)=a{{\left( x-\alpha \right)}^{2}},where\,\alpha =\frac{-b}{2a}.\)

(iii) If D > 0 & if

  • A > 0, then f(x) > 0 \(\forall\) X ϵ R and the expression will be cross multiplied and the sign of the inequality will not change.
  • A < 0, then f(x) < 0 \(\forall\) X ϵ R and the expression will be cross multiplied and the sign of the inequality will change.
  • If the expression (say ‘f’) is cancelled from the same side of the inequality, then cancel it and write f ≠ 0 e.g.,

(i) \(\frac{\left( x-1 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-5 \right)}>1\,\,\,\Rightarrow \,\,\,\frac{\left( x-3 \right)}{\left( x-5 \right)}>1\,\,\,\,\,\,\,\,\,\,\,iff\,\,x-2\ne 0\)

(ii) \(\frac{{{\left( x-5 \right)}^{2}}\left( x-8 \right)}{\left( x-5 \right)}\ge 0\,\,\,\Rightarrow \,\,\,\left( x-5 \right)\,\left( x-8 \right)\ge 0\,\,\,\,\,\,\,\,\,\,\,iff\,\,x-5\ne 0\)

Types of Inequalities

Type I: Inequalities involving non-repeating linear factors

1st condition\(\left. \begin{matrix} \left( x-1 \right)>0\Rightarrow x>1 \\ \left( x-2 \right)>0\Rightarrow x>2 \\ \end{matrix} \right\}x\ge 2\).

2nd condition \(\left. \begin{matrix} x-1<0\Rightarrow x<1 \\ x-2<0\Rightarrow x>2 \\ \end{matrix} \right\}x\le 1\)

∴ \(\,x\,\varepsilon \,(-\infty ,1]\cup [2,\infty )\)

Illustration: (x – 3) (x + 1) \(\left( x-\frac{12}{7} \right)<0,\) find range of x

Sol: Comparing all brackets separately with 0, we can find the range of values for x.

\(x<-1\,and\,\frac{12}{7}\,<\,x\,<\,3;\,∴ \,x\,\varepsilon \,\left( -\infty ,-1 \right)\cup \left( \frac{12}{7},3 \right)\)

Type II: Inequalities involving repeating linear factors

\({{\left( x-1 \right)}^{2}}{{\left( x+2 \right)}^{3}}\left( x-3 \right)\le 0\) \(\Rightarrow \,{{\left( x+1 \right)}^{2}}{{\left( x+2 \right)}^{2}}\left( x+2 \right)\left( x-3 \right)\le 0\) \(\Rightarrow \,\left( x+2 \right)\left( x-3 \right)\le 0;x\,\varepsilon [-2,3]\)

Illustration: Find the greatest integer satisfying the equation.

\({{\left( x+1 \right)}^{101}}{{\left( x-3 \right)}^{2}}{{\left( x-5 \right)}^{11}}{{\left( x-4 \right)}^{200}}{{\left( x-2 \right)}^{555}}<0\)

Sol: Comparing all brackets separately with 0, we can find the greatest integer.

The inequality \({{\left\{ -\left( x-2 \right) \right\}}^{2}}-\left( x-2 \right)-2=0\)

Þ (x + 1) (x – 5) (x – 2) < 0

x = –1 , 3, 5, 4, 2

\(x\in \left( -\infty ,-1 \right)\cup (2,3)\cup (3,4)\cup (4,5).\)

Type III: Inequalities expressed in rational form.

Illustration: \(\frac{\left( x-1 \right)\left( x+2 \right)}{\left( x+3 \right)\left( x-4 \right)}\ge 0\)

Sol: If \(\frac{\left( x+a \right)\left( x+b \right)}{\left( x+c \right)\left( x-d \right)}\ge 0\,then\,\left( x+c \right)\left( x+d \right)\ne 0,\,and\,\left( x+a \right)\left( x+b \right)=0\)

Hence, x ≠ –3, 4 &     x = 1, –2;         \(x\,\varepsilon \left( -\infty ,-3 \right)\cup [-2,1]\cup (4,\infty )\)

Illustration: \(\frac{{{x}^{2}}(x+1)}{{{(x-3)}^{3}}}<0\)

Sol: Similar to the illustration above.

\(\frac{x+1}{x-3}<0\,\,\,x\ne 3,\,-1,0;\,\,\,\,\,x\,\in \,\left( -1,0 \right)\cup (0,3)\)

Illustration: \(\frac{{{x}^{2}}-1}{{{x}^{2}}-7x+12}\ge 1\)

Sol: First reduce the given inequalities in rational form and then solve it in the manner similar to the illustration above.

\(\frac{{{x}^{2}}-1}{\left( x-4 \right)\left( x-3 \right)}\ge 1\) \(∴ \,\frac{\left( x+1 \right)\left( x-1 \right)}{\left( x-4 \right)\left( x-3 \right)}\ge 1\,\,\,\,\Rightarrow \frac{{{x}^{2}}-1}{{{x}^{2}}-7x+12}-1\ge 0\) \(∴ \,\frac{x2-1-x2+7x-12}{\left( x-4 \right)\left( x-3 \right)}\ge 0\,\,\,\,\,∴ \frac{7x-13}{\left( x-4 \right)\left( x-3 \right)}\ge 0\) \(∴ \,x\ne 3,4;\,\,\,\,\,\,\,\,\,\,x\varepsilon \left[ \frac{13}{7},3 \right)\cup (4,\infty )\)

Type IV: Double inequality

Illustration \(1<\frac{3{{x}^{2}}-7x+8}{{{x}^{2}}+1}\le 2\)

Sol: Here 3x2 – 7x + 8 > x2 + 1 therefore if D < 0 & if a > 0, then f(x) > 0 and always positive for all real x.

3x2 – 7x + 8 > x2 + 1 Þ 2x2 – 7x + 7 > 0

D = b2 – 4ac = 49 – 56 = –7

\(∴ \,\,\,D<0\,\,\And \,\,a>0\,\,\,\,\,\,\,\,∴ \,always\,positive\,for\,all\,real\,x\)

3x2 – 7x + 8 < 2x2 + 2 Þ x2 – 7x + 6 < 0 Þ \(\left( x-1 \right)\left( x-6 \right)\le 0\) \(x\in [1,6];x\varepsilon \,[1,6]\cap R\)

Type V: Inequalities involving biquadrate expression

Illustration \(\left( {{x}^{2}}+3x+1 \right)\left( {{x}^{2}}+3x-3 \right)\ge 5\)

Sol: Using x2 + 3x = y, we can solve this problem

Let x2 + 3x = y                                     \(∴ \left( y+1 \right)\left( y-3 \right)\ge 5\)

y2 – 2y – 8 > 0                         \(∴ \left( y-4 \right)\left( y+2 \right)\ge 0\) \(∴ (x+4)(x-1)(x+2)(x+1)\ge 0\Rightarrow x\in \left( -\infty ,-4 \right]\cup \left[ -2,-1 \right]\cup \left[ 1,\infty \right)\)

Some Basic Properties of Inequality

Intervals

Given E(x) = (x – a)(x – b)(x – c)(x – d) > 0

To find the solution set of the above inequality we have to check the intervals in which E(x) is greater/less than zero.

  • Closed Interval: The set of all value of x, which lies between a & b and is also equal to a & b is known as a closed interval, i.e. if a < x < b then it is denoted by x ϵ [a, b].
  • Open-Closed Interval: The set of all values of x, which lies between a & b, equal to b, but not equal to a is known as an open-closed interval, i.e. if a < x < b then it is denoted by x ϵ (a, b)
  • Open – Closed Interval: The set of all values of x, which lies between a & b, equal to b, but not equal to a is known as an open-closed interval, i.e. if a < x < b then it is denoted by x ϵ (a, b].
  • Closed-open Interval: The set of all values of x, which lies between a & b, equal to a but not equal to b is called a closed-open interval, ie. if a < x < b, then it is denoted by c ϵ [a, b)

Note:

\((i)x\ge a\Rightarrow [a,\infty )\) \((ii)x>a\Rightarrow [a,\infty )\) \((iii)x\le a\Rightarrow [-\infty ,a)\) \((iv)x<a\Rightarrow (-\infty ,a)\)

Properties

(a) In an inequality, any number can be added or subtracted from both sides of inequality.

(b) Terms can be shifted from one side to the other side of the inequality. The sign of inequality does not change.

(c) If we multiply both sides of the inequality by a non zero positive number, then the sign of inequality does not change.

But if we multiply both sides of the inequality by a non-zero negative number then the sign of the inequality does get changed.

(d) In the inequality, if the sign of an expression is not known then it cannot be cross multiplied. Similarly, without

knowing the sign of an expression, division is not possible.

(i) \(\frac{x-2}{x-5}>1\Rightarrow x-2>x-5\) (Not valid because we don’t know the sign of the expression)

(ii) \(\frac{x-2}{{{\left( x-5 \right)}^{2}}}>1\Rightarrow \left( x-2 \right)>{{\left( x-5 \right)}^{2}}\) (valid because (x – 4)2 is always positive)

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.

Quadratic Equations IIT JEE

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

Using Quadratic Formula:

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.

Quadratic Equations IIT JEE

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

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

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