Algebra (Olympiad)hardFree

Algebra (Olympiad): Real Number Find Value

JEE Maths question with a full step-by-step solution.

Question
If xx is a real number such that
x+(x+1)(x+2)+(x+2)(x+3)+(x+3)(x+1)=4,x + \sqrt{(x+1)(x+2)} + \sqrt{(x+2)(x+3)} + \sqrt{(x+3)(x+1)} = 4,
 find the value of PP, where x+37=P840x + \dfrac{3}{7} = \dfrac{P}{840}.
Solution
Answer: 49
Step 1: Absorb the constant into xx. Adding 11 to both sides,
(x+1)+(x+1)(x+2)+(x+2)(x+3)+(x+3)(x+1)=5.(x+1) + \sqrt{(x+1)(x+2)} + \sqrt{(x+2)(x+3)} + \sqrt{(x+3)(x+1)} = 5.
 Step 2: Factor the left side. For x≥−1x \ge -1 each of x+1\sqrt{x+1}, x+2\sqrt{x+2}, x+3\sqrt{x+3} is real. Grouping,
x+1(x+1+x+2)+x+3(x+1+x+2),\sqrt{x+1}\big(\sqrt{x+1} + \sqrt{x+2}\big) + \sqrt{x+3}\big(\sqrt{x+1} + \sqrt{x+2}\big),
 which factors as
(x+1+x+3)(x+1+x+2)=5.\big(\sqrt{x+1} + \sqrt{x+3}\big)\big(\sqrt{x+1} + \sqrt{x+2}\big) = 5.
 Step 3: Isolate the first factor. Dividing by (x+1+x+2)\big(\sqrt{x+1} + \sqrt{x+2}\big),
x+1+x+3=5x+1+x+2.\sqrt{x+1} + \sqrt{x+3} = \frac{5}{\sqrt{x+1} + \sqrt{x+2}}.
 Rationalize the right side by multiplying numerator and denominator by (x+2−x+1)\big(\sqrt{x+2} - \sqrt{x+1}\big). Since (x+2)−(x+1)=1(x+2) - (x+1) = 1,
5x+1+x+2=5(x+2−x+1).\frac{5}{\sqrt{x+1} + \sqrt{x+2}} = 5\big(\sqrt{x+2} - \sqrt{x+1}\big).
 Therefore
x+1+x+3=5x+2−5x+1.\sqrt{x+1} + \sqrt{x+3} = 5\sqrt{x+2} - 5\sqrt{x+1}.
 Step 4: Collect like terms.
6x+1+x+3=5x+2.6\sqrt{x+1} + \sqrt{x+3} = 5\sqrt{x+2}.
 Step 5: Square both sides.
36(x+1)+(x+3)+12(x+1)(x+3)=25(x+2).36(x+1) + (x+3) + 12\sqrt{(x+1)(x+3)} = 25(x+2).
 Combine the non-radical terms, using (x+1)(x+3)=x2+4x+3(x+1)(x+3) = x^2 + 4x + 3:
37x+39+12x2+4x+3=25x+50,37x + 39 + 12\sqrt{x^2 + 4x + 3} = 25x + 50,
 so
12x2+4x+3=−12x+11.12\sqrt{x^2 + 4x + 3} = -12x + 11.
 Step 6: Square again.
144(x2+4x+3)=(−12x+11)2=144x2−264x+121.144(x^2 + 4x + 3) = (-12x + 11)^2 = 144x^2 - 264x + 121.
 Expanding the left side and cancelling 144x2144x^2,
576x+432=−264x+121.576x + 432 = -264x + 121.
 Step 7: Solve for xx.
(576+264)x=121−432  ⟹  840x=−311  ⟹  x=−311840.(576 + 264)x = 121 - 432 \implies 840x = -311 \implies x = -\frac{311}{840}.
 Step 8: Verify in the original equation. With x=−311840x = -\dfrac{311}{840},
x+1=529840=232840,x+2=1369840=372840,x+3=2209840=472840.x + 1 = \frac{529}{840} = \frac{23^2}{840}, \qquad x + 2 = \frac{1369}{840} = \frac{37^2}{840}, \qquad x + 3 = \frac{2209}{840} = \frac{47^2}{840}.
 Hence
(x+1)(x+2)=23⋅37840=851840,(x+2)(x+3)=37⋅47840=1739840\sqrt{(x+1)(x+2)} = \frac{23 \cdot 37}{840} = \frac{851}{840}, \quad \sqrt{(x+2)(x+3)} = \frac{37 \cdot 47}{840} = \frac{1739}{840}
,
(x+3)(x+1)=47⋅23840=1081840,\sqrt{(x+3)(x+1)} = \frac{47 \cdot 23}{840} = \frac{1081}{840},
 and the left side becomes
−311840+851+1739+1081840=−311+3671840=3360840=4,-\frac{311}{840} + \frac{851 + 1739 + 1081}{840} = \frac{-311 + 3671}{840} = \frac{3360}{840} = 4,
 which confirms the value. Step 9: Compute PP. Since 37=360840\dfrac{3}{7} = \dfrac{360}{840},
x+37=−311840+360840=49840=P840  ⟹  P=49.x + \frac{3}{7} = -\frac{311}{840} + \frac{360}{840} = \frac{49}{840} = \frac{P}{840} \implies P = 49.
 Answer: 49
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