Twist and Turns!

Algebra Level 3

5 x 2 + 5 y 2 − 8 x y − 2 x − 4 y + 5 = 0 5x^2 + 5y^2 - 8xy- 2x - 4y + 5 = 0

Is the above equation satisfied by any pair of real numbers x x and y y ?

Yes No

Achal Jain by Achal Jain , 19, India

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1 solution

Rearranging the terms, we get 5 x 2 − ( 8 y + 2 ) x + ( 5 y 2 + 5 ) = 0 5x^2-(8y+2)x+(5y^2+5)=0 Solving in terms of x x , we have x = 8 y + 2 ± ( 8 y + 2 ) 2 − 4 ⋅ 5 ⋅ ( 5 y 2 + 5 ) 10 x=\frac{8y+2\pm\sqrt{(8y+2)^2-4\cdot5\cdot(5y^2+5)}}{10} In order for x x to be real, we know that ( 8 y + 2 ) 2 − 4 ⋅ 5 ⋅ ( 5 y 2 + 5 ) ≥ 0 (8y+2)^2-4\cdot5\cdot(5y^2+5)\geq0 − 9 y 2 + 8 y − 24 ≥ 0 -9y^2+8y-24\geq0 However, plotting the graph, we realise that it is always negative, so x x is always complex, and the answer is NO

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The correct method i would have to say :-)

Achal Jain - 5 years, 4 months ago

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