Theory of Intelligence
11TH•Mathematics•Theory Of Quadratic Functions•Question 4

Step-by-Step Solution

Solve the following absolute value quadratic equations and inequalities:

Step-by-Step Solution

Step 1: Use the Disjunction Property of Absolute Values

For any positive constant c>0c > 0, the inequality ∣A∣>c|A| > c splits into two disjoint possibilities:

A>corA<−cA > c \quad \text{or} \quad A < -c

Substituting A=x2−3x+2A = x^2 - 3x + 2 and c=4c = 4:

x2−3x+2>4orx2−3x+2<−4x^2 - 3x + 2 > 4 \quad \text{or} \quad x^2 - 3x + 2 < -4

Step 2: Solve Branch 1 (>4> 4)

x2−3x+2>4x2−3x−2>0\begin{aligned} x^2 - 3x + 2 &> 4 \\ x^2 - 3x - 2 &> 0 \end{aligned}

Find the roots of the corresponding equation x2−3x−2=0x^2 - 3x - 2 = 0:

x=−(−3)±(−3)2−4(1)(−2)2(1)=3±9+82=3±172x = \frac{-(-3) \pm\sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} = \frac{3 \pm\sqrt{9 + 8}}{2} = \frac{3 \pm\sqrt{17}}{2}

Numerically: 3−4.1232≈−0.562\frac{3 - 4.123}{2} \approx -0.562 and 3+4.1232≈3.562\frac{3 + 4.123}{2} \approx 3.562.

Because the leading coefficient is positive (a=1>0a = 1 > 0), the upward parabola is strictly positive outside the roots:

x∈(−∞,3−172)∪(3+172,∞)x \in \left(-\infty, \frac{3 - \sqrt{17}}{2}\right) \cup \left(\frac{3 + \sqrt{17}}{2}, \infty\right)

Step 3: Solve Branch 2 (<−4< -4)

x2−3x+2<−4x2−3x+6<0\begin{aligned} x^2 - 3x + 2 &< -4 \\ x^2 - 3x + 6 &< 0 \end{aligned}

Compute the discriminant of x2−3x+6x^2 - 3x + 6:

Δ=(−3)2−4(1)(6)=9−24=−15<0\Delta = (-3)^2 - 4(1)(6) = 9 - 24 = -15 < 0

Since Δ<0\Delta < 0 and a=1>0a = 1 > 0, the parabola y=x2−3x+6y = x^2 - 3x + 6 lies entirely above the xx-axis and is strictly positive for all x∈Rx \in \mathbb{R}. It can never be negative. Thus, Branch 2 produces no solutions (∅\emptyset).

Step 4: Union of Branches

The complete solution set is given entirely by Branch 1:

x∈(−∞,3−172)∪(3+172,∞)x \in \left(-\infty, \frac{3 - \sqrt{17}}{2}\right) \cup \left(\frac{3 + \sqrt{17}}{2}, \infty\right)

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