Basics & LogarithmsmediumFree

Basics & Logarithms: Possible Solutions System Equal

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression 11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}} is
Basics & Logarithms · easy
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
Basics & Logarithms · easy
Number of natural numbers for which the number log⁡4−x(x2−14x+45)\log_{4-x}(x^2-14x+45) is defined is:
Basics & Logarithms · easy
The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_{e}x)^{2}+3}=x^{8}, x>0x>0, is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath