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 (English)

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.

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