Check whether the following are : (i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x) (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5) (v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Step-by-Step Solution
Key Concept: Expand both sides of each equation, bring all terms to one side, simplify and factor (or use the quadratic formula) to obtain a polynomial equation. If the resulting polynomial reduces to 0 for all values of \(x\) it is an identity; if it reduces to a non‑zero constant it is a contradiction; otherwise it is a conditional equation whose solutions are obtained by solving the resulting polynomial.
i) \((x+1)^2 = 2(x-3)\)
Expand: \(x^2+2x+1 = 2x-6\)
Bring all terms to left: \(x^2+2x+1-2x+6 = 0 \Rightarrow x^2+7=0\)
\(x^2 = -7\) has no real solution ⇒ contradiction.
ii) \(x^2-2x = (-2)(3-x)\)
RHS = \(-6+2x\)
Equation: \(x^2-2x = -6+2x\)
\(x^2-4x+6 = 0\)
Discriminant \(D = 16-24 = -8 < 0\) ⇒ no real solution ⇒ contradiction.
iii) \((x-2)(x+1) = (x-1)(x+3)\)
Expand: \(x^2 - x - 2 = x^2 + 2x - 3\)
Simplify: \(-x-2 = 2x-3\)
\(-3x +1 = 0 \Rightarrow x = \frac{1}{3}\)
Hence a conditional equation with solution \(x=\frac13\).
iv) \((x-3)(2x+1) = x(x+5)\)
Expand: \(2x^2 -5x -3 = x^2 +5x\)
\(x^2 -10x -3 = 0\)
Using quadratic formula: \(x = \frac{10 \pm \sqrt{100+12}}{2}=5 \pm 2\sqrt{7}\)
Conditional equation with two real solutions.
v) \((2x-1)(x-3) = (x+5)(x-1)\)
Expand: \(2x^2 -7x +3 = x^2 +4x -5\)
\(x^2 -11x +8 = 0\)
Solutions: \(x = \frac{11 \pm \sqrt{89}}{2}\)
Conditional equation with two real solutions.
vi) \(x^2 +3x +1 = (x-2)^2\)
RHS = \(x^2 -4x +4\)
Cancel \(x^2\): \(3x +1 = -4x +4\)
\(7x = 3 \Rightarrow x = \frac{3}{7}\)
Conditional equation with single solution.
vii) \((x+2)^3 = 2x(x^2-1)\)
Expand: \(x^3 +6x^2 +12x +8 = 2x^3 -2x\)
Bring to left: \(-x^3 +6x^2 +14x +8 = 0\)
Multiply by \(-1\): \(x^3 -6x^2 -14x -8 = 0\)
The cubic does not factor over rational numbers; it has one real root (approximately \(x \approx 7.9\)) and two complex conjugate roots. Hence it is a conditional equation (real solution \(x \approx 7.9\)).
viii) \(x^3 -4x^2 -x +1 = (x-2)^3\)
RHS = \(x^3 -6x^2 +12x -8\)
Cancel \(x^3\): \(-4x^2 -x +1 = -6x^2 +12x -8\)
Bring all to left: \(2x^2 -13x +9 = 0\)
Solutions: \(x = \frac{13 \pm \sqrt{97}}{2}\)
Conditional equation with two real solutions.
Correct Answer: i) Contradiction (no real solution)\nii) Contradiction (no real solution)\niii) Conditional – solution \(x = \frac{1}{3}\)\niv) Conditional – solutions \(x = 5 \pm 2\sqrt{7}\)\nv) Conditional – solutions \(x = \frac{11 \pm \sqrt{89}}{2}\)\nvi) Conditional – solution \(x = \frac{3}{7}\)\nvii) Conditional – one real solution \(x \approx 7.9\) (other two roots are complex)\nviii) Conditional – solutions \(x = \frac{13 \pm \sqrt{97}}{2}\)