For a ≤ 0, determine all real roots of...

MathematicsSolution of Quadratic EquationsJEE Advanced 1986Moderate

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}

Visualized Solution (Hindi)

Identifying the Pivot Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x = a

Given equation: $x^{2} - 2 a ∣ x - a ∣ - 3 a^{2} = 0$ with $a \leq 0$ .
The critical point for the modulus $∣ x - a ∣$ is $x = a$ .
We must analyze the equation in two domains: $x \geq a$ and $x < a$ .

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x \geq a

For $x \geq a$ , $∣ x - a ∣ = x - a$ .
Substitute into the equation: $x^{2} - 2 a (x - a) - 3 a^{2} = 0$ .
Expand the terms: $x^{2} - 2 a x + 2 a^{2} - 3 a^{2} = 0$ .

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

Simplified Quadratic: $x^{2} - 2 a x - a^{2} = 0$ .
Using Quadratic Formula: $x = \frac{- ( - 2 a ) \pm ( - 2 a ) ^{2} - 4 ( 1 ) ( - a ^{2} )}{2 ( 1 )}$ .
Calculation: $x = \frac{2 a \pm 4 a ^{2} + 4 a ^{2}}{2} = \frac{2 a \pm 2 a 2}{2}$ .
Roots for Case 1: $x = a \pm a 2$ .

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x < a

For $x < a$ , $∣ x - a ∣ = - (x - a) = a - x$ .
Substitute into the equation: $x^{2} - 2 a (a - x) - 3 a^{2} = 0$ .
Expand the terms: $x^{2} - 2 a^{2} + 2 a x - 3 a^{2} = 0$ .

Solving Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x^{2} + 2 a x - 5 a^{2} = 0

Simplified Quadratic: $x^{2} + 2 a x - 5 a^{2} = 0$ .
Using Quadratic Formula: $x = \frac{- ( 2 a ) \pm ( 2 a ) ^{2} - 4 ( 1 ) ( - 5 a ^{2} )}{2 ( 1 )}$ .
Calculation: $x = \frac{- 2 a \pm 4 a ^{2} + 20 a ^{2}}{2} = \frac{- 2 a \pm 2 a 6}{2}$ .
Roots for Case 2: $x = - a \pm a 6$ .

Summary of Real Roots

The complete set of real roots is obtained by combining both cases.
Final Roots: ${a \pm a 2, - a \pm a 6}$ .
Key Takeaway: Always split absolute value equations at their critical points to simplify the problem into standard algebraic forms.

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Solve the following equation for $x : 2 lo g_{x} a + lo g_{a x} a + 3 lo g_{a^{2} x} a = 0, a > 0, a \neq = 0$

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Let $a, b, c$ be real numbers, $a \neq = 0$ . If $α$ is a root of $a^{2} x^{2} + b x + c = 0$ . $β$ is the root of $a^{2} x^{2} - b x - c = 0$ and $0 < α < β$ , then the equation $a^{2} x^{2} + 2 b x + 2 c = 0$ has a root $γ$ that always satisfies

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

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Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

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

Visualized Solution (Hindi)

Identifying the Pivot Point $x = a$

Case 1: $x \geq a$

Solving Case 1: $x^{2} - 2 a x - a^{2} = 0$

Case 2: $x < a$

Solving Case 2: $x^{2} + 2 a x - 5 a^{2} = 0$

Summary of Real Roots

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a≤0, determine all real roots of the equation x2−2a∣x−a∣−3a2=0

Visualized Solution (Hindi)

Identifying the Pivot Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=a

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥a

Solving Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2−2ax−a2=0

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<a

Solving Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+2ax−5a2=0

Summary of Real Roots

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

Identifying the Pivot Point $x = a$

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Solving Case 1: $x^{2} - 2 a x - a^{2} = 0$

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