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

Step-by-Step Solution

Solve the following absolute value quadratic equations and inequalities:

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Step-by-Step Solution

Step 1: Split into Two Disjoint Cases

For any real expression $|A| = c$ (with $c > 0$ ), we have $A = c$ or $A = -c$ :

x^2 - 6x + 8 = 4 \quad \text{or} \quad x^2 - 6x + 8 = -4

Step 2: Solve Case 1 ( $+4$ )

\begin{aligned} x^2 - 6x + 8 &= 4 \\ x^2 - 6x + 4 &= 0 \end{aligned}

Using the quadratic formula:

\begin{aligned} x &= \frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(4)}}{2(1)} \\ &= \frac{6 \pm\sqrt{36 - 16}}{2} \\ &= \frac{6 \pm\sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} \\ &= 3 \pm\sqrt{5} \end{aligned}

Numerically: $x \approx 3 - 2.236 = 0.764$ and $x \approx 3 + 2.236 = 5.236$ .

Step 3: Solve Case 2 ( $-4$ )

\begin{aligned} x^2 - 6x + 8 &= -4 \\ x^2 - 6x + 12 &= 0 \end{aligned}

Compute the discriminant:

\Delta = (-6)^2 - 4(1)(12) = 36 - 48 = -12 < 0

Since the discriminant is strictly negative, Case 2 produces no real solutions.

Step 4: Summary of Real Roots

Combining both cases, the real solutions are precisely:

x = 3 \pm\sqrt{5}

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