Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

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

Question
If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are all possible solutions of the system x2+y2+6x+2y=0x^2 + y^2 + 6x + 2y = 0 and x+y+8=0x + y + 8 = 0, then x12+x22+y12+y22x_1^2 + x_2^2 + y_1^2 + y_2^2 is equal to:
A6868
B3232
C3838
D7272correct
Solution
Step 1: Substitute the linear relation From x+y+8=0x+y+8 = 0, y=−8−xy = -8-x. Substituting into the first equation:
x2+(−8−x)2+6x+2(−8−x)=0x^2 + (-8-x)^2 + 6x + 2(-8-x) = 0
 Step 2: Simplify to a quadratic in xx
2x2+20x+48=0  ⟹  x2+10x+24=0  ⟹  (x+4)(x+6)=02x^2 + 20x + 48 = 0 \implies x^2 + 10x + 24 = 0 \implies (x+4)(x+6) = 0
x=−4  ⟹  y=−4,x=−6  ⟹  y=−2x = -4 \implies y = -4, \qquad x = -6 \implies y = -2
 Step 3: Evaluate the required sum
x12+x22+y12+y22=(−4)2+(−6)2+(−4)2+(−2)2=16+36+16+4=72x_1^2 + x_2^2 + y_1^2 + y_2^2 = (-4)^2 + (-6)^2 + (-4)^2 + (-2)^2 = 16+36+16+4 = 72
 Answer: (4)
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