Find all real values of x which satisfy...

MathematicsSolution of Quadratic EquationsJEE Advanced 1983Easy

Find all real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 2 x - 4 \leq 0$

Answer:

x \in [1 - 5, 1) \cup (2, 1 + 5]

Visualized Solution (Hindi)

Introduction to the System

Given System:
1. $x^{2} - 3 x + 2 > 0$
2. $x^{2} - 2 x - 4 \leq 0$
Goal: Find the intersection of the solution sets of both inequalities.

Factoring the First Inequality

Inequality 1: $x^{2} - 3 x + 2 > 0$
Factorizing the quadratic:
Find numbers that sum to $- 3$ and multiply to $2$ .
$(x - 1) (x - 2) > 0$

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x - 1) (x - 2) > 0

Critical points: $x = 1, x = 2$
Using the sign scheme:
Positive in $(- \infty, 1) \cup (2, \infty)$
Solution 1: $x \in (- \infty, 1) \cup (2, \infty)$

Visualizing Inequality 1

Visualizing Inequality 1:
Region: $x < 1$ or $x > 2$
Represented by blue rays on the number line.

Analyzing the Second Inequality

Inequality 2: $x^{2} - 2 x - 4 \leq 0$
This does not factorize easily with integers.
We will use the quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Finding Roots via Quadratic Formula

For $x^{2} - 2 x - 4 = 0$ :
$a = 1, b = - 2, c = - 4$
$x = \frac{2 \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - 4 )}{2 ( 1 )}$
$x = \frac{2 \pm 4 + 16}{2} = \frac{2 \pm 20}{2}$
$x = \frac{2 \pm 2 5}{2} = 1 \pm 5$

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x^{2} - 2 x - 4 \leq 0

Roots: $x_{1} = 1 - 5, x_{2} = 1 + 5$
Since the parabola opens upwards ( $a > 0$ ), the expression is $\leq 0$ between the roots.
Solution 2: $x \in [1 - 5, 1 + 5]$

Visualizing Inequality 2

Visualizing Inequality 2:
Region: $1 - 5 \leq x \leq 1 + 5$
Represented by the green segment on the number line.

Finding the Intersection

Finding Intersection:
Region 1: $(- \infty, 1) \cup (2, \infty)$
Region 2: $[1 - 5, 1 + 5]$
Overlap occurs in two parts:
1. $[1 - 5, 1)$
2. $(2, 1 + 5]$

Final Solution and Summary

Final Answer:
x \in [1 - \sqrt{5}, 1) \cup (2, 1 + \sqrt{5}]
Key Takeaways:
* Factorize simple quadratics first.
* Use the quadratic formula for non-integer roots.
* Use a number line to find the intersection of multiple sets.

00:00 / 00:00

Conceptually Similar Problems

MathematicsSolution of Quadratic EquationsJEE Advanced 2002Easy

View in:English Hindi

The set of all real numbers $x$ for which $x^{2} - ∣ x + 2∣ + x > 0$ , is

(A)

(- \infty, - 2) \cup (2, \infty)

(B)

(- \infty, - 2) \cup (2, \infty)

(C)

(- \infty, - 1) \cup (1, \infty)

(D)

(2, \infty)

Answer:B

MathematicsLinear InequalitiesJEE Advanced 1987Moderate

View in:English Hindi

Find the set of all $x$ for which $\frac{2 x}{2 x ^{2} + 5 x + 2} > \frac{1}{x + 1}$

Answer:

(- 2, - 1) \cup (- 2/3, - 1/2)

MathematicsTrigonometric Ratios and IdentitiesJEE Advanced 1994Moderate

View in:English Hindi

Let $2 sin^{2} x + 3 sin x - 2 > 0$ and $x^{2} - x - 2 < 0$ ( $x$ is measured in radians). Then $x$ lies in the interval

(A)

(π /6, 5 π /6)

(B)

(- 1, 5 π /6)

(C)

(- 1, 2)

(D)

(π /6, 2)

Answer:D

MathematicsLinear InequalitiesJEE Advanced 1980Easy

View in:English Hindi

Let $y = \frac{( x + 1 ) ( x - 3 )}{( x - 2 )}$ . Find all the real values of $x$ for which $y$ takes real values.

Answer:

x \in [- 1, 2) \cup [3, \infty)

MathematicsSolution of Quadratic EquationsJEE Advanced 1986Moderate

View in:English Hindi

For $a \leq 0$ , determine all real roots of the equation $x^{2} - 2 a ∣ x - a ∣ - 3 a^{2} = 0$

Answer:

{a \pm a 2, - a \pm a 6}

MathematicsProperties of Modulus FunctionJEE Advanced 1983Moderate

View in:English Hindi

If $x$ satisfies $∣ x - 1∣ + ∣ x - 2∣ + ∣ x - 3∣ \geq 6$ , then

(A)

0 \leq x \leq 4

(B)

x \leq - 2

x \geq 4

(C)

x \leq 0

x \geq 4

(D)

None of these

Answer:C

MathematicsMaximum and Minimum Values of Quadratic ExpressionsJEE Advanced 2005Moderate

View in:English Hindi

Find the range of values of $t$ for which $2 sin t = \frac{1 - 2 x + 5 x ^{2}}{3 x ^{2} - 2 x - 1}$ , $t \in [- π /2, π /2]$ .

Answer:

[- \frac{π}{2}, - \frac{π}{10}] \cup [\frac{3 π}{10}, \frac{π}{2}]

MathematicsLocation of RootsJEE Main 2006Easy

View in:English Hindi

All the values of $m$ for which both roots of the equation $x^{2} - 2 m x + m^{2} - 1 = 0$ are greater than - 2 but less than 4, lie in the interval

(A)

- 2 < m < 0

(B)

m > 3

(C)

- 1 < m < 3

(D)

1 < m < 4

Answer:C

MathematicsLocation of RootsJEE Advanced 1999Moderate

View in:English Hindi

If the roots of the equation $x^{2} - 2 a x + a^{2} + a - 3 = 0$ are real and less than 3, then

(A)

a < 2

(B)

2 \leq a \leq 3

(C)

3 < a \leq 4

(D)

a > 4

Answer:A

MathematicsStandard and General Equation of a CircleJEE Advanced 1978Moderate

View in:English Hindi