Algebra_1

Algebra Level 2

If x + y + z = 0 x + y + z = 0 , then what is the value of 3 y 2 + x 2 + z 2 2 y 2 − x z \dfrac{3y^{2} + x^{2} + z^{2}}{2y^{2} - xz} ?


The answer is 2.

Mehakdeep Kaur by Mehakdeep Kaur , 26, India

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2 solutions

X X
Apr 15, 2018

y 2 = ( x + z ) 2 \displaystyle y^2=(x+z)^{2} ,so 3 y 2 + x 2 + z 2 2 y 2 − x z = 3 ( x + z ) 2 + x 2 + z 2 2 ( x + z ) 2 − x z = 4 x 2 + 6 x z + 4 z 2 2 x 2 + 3 x z + 2 z 2 = 2 \displaystyle \frac{3y^{2} + x^{2} + z^{2}}{2y^{2} - xz}=\frac{3(x+z)^{2} + x^{2} + z^{2}}{2(x+z)^{2} - xz}=\frac{4x^{2} + 6xz + 4z^{2}}{2x^2+ 3xz+2z^2}=2

1 Helpful 0 Interesting 0 Brilliant 0 Confused
A I
Apr 15, 2018

All I did was to substitute z with x and y and found that the numerator is twice the denominator.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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