Basics & LogarithmsmediumPYQ · JEE Main · 8 Apr 2025 · Shift 2 (Afternoon)Free

Basics & Logarithms: Sum Squares Roots Squares Roots (JEE Main 2025)

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

Question
The sum of the squares of the roots of ∣x−2∣2+∣x−2∣−2=0|x-2|^{2}+|x-2|-2=0 and the squares of the roots of x2−2∣x−3∣−5=0x^{2}-2|x-3|-5=0 is
A2424
B3636correct
C3030
D2626
Solution
Step 1: Solve the first equation ∣x−2∣2+∣x−2∣−2=0|x-2|^{2}+|x-2|-2=0. Let u=∣x−2∣u=|x-2|. Then u2+u−2=0⇒(u+2)(u−1)=0u^{2}+u-2=0\Rightarrow(u+2)(u-1)=0. So u=1u=1 or u=−2u=-2 (rejected since ∣x−2∣≥0|x-2|\ge 0). Step 2: From ∣x−2∣=1|x-2|=1: x−2=±1x-2=\pm 1, giving x=1x=1 or x=3x=3. Sum of squares: 12+32=1+9=101^{2}+3^{2}=1+9=10. Step 3: Solve the second equation x2−2∣x−3∣−5=0x^{2}-2|x-3|-5=0. Case I: x≥3x\ge 3. Then ∣x−3∣=x−3|x-3|=x-3:
x2−2(x−3)−5=0  ⇒  x2−2x+1=0  ⇒  (x−1)2=0  ⇒  x=1x^{2}-2(x-3)-5=0\;\Rightarrow\;x^{2}-2x+1=0\;\Rightarrow\;(x-1)^{2}=0\;\Rightarrow\;x=1
 But x=1<3x=1<3, so rejected. Step 4: Case II: x<3x<3. Then ∣x−3∣=−(x−3)=3−x|x-3|=-(x-3)=3-x:
x2−2(3−x)−5=0  ⇒  x2+2x−11=0  ⇒  (x+1)2=12x^{2}-2(3-x)-5=0\;\Rightarrow\;x^{2}+2x-11=0\;\Rightarrow\;(x+1)^{2}=12
x=−1±23x=-1\pm 2\sqrt{3}
 Both values lie in x<3x<3 (since −1+23≈2.46<3-1+2\sqrt{3}\approx 2.46<3), accepted. Step 5: Sum of squares of these roots:
(−1+23)2+(−1−23)2=2(1+12)=26(-1+2\sqrt{3})^{2}+(-1-2\sqrt{3})^{2}=2(1+12)=26
 Step 6: Total sum of squares: 10+26=3610+26=36. Answer: (2) 3636
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