Help with solution to 2 variable quadratic equation. Could

anadyrskia0g5

anadyrskia0g5

Answered question

2022-03-26

Help with solution to 2 variable quadratic equation.
Could someone help with an explanation on how to treat the equation below, please? A solution I have read uses the discriminant to find the range i.e.
920y216y>or=0. To me this means treating the variable Y as a constant , is that "acceptable" ? does anyone have a reference for these equations ?
yx2+5y3x+4=0
rewrite as yx23x+(5y+4)=0

Answer & Explanation

exinnaemekswr1k

exinnaemekswr1k

Beginner2022-03-27Added 10 answers

Yes, I take from yx23x+(5y+4)0≈∼(1)
A quadratic f(x)=Ax2+Bx+C is positive definite for all real values of x if B24AC and A>0 So (1) will hold for all real values of x if y>0 and B24AC. So we get
94y(5y+4)20y2+6y90
(x4+6110)(x46110)0
So y4+6110ory46110
But due to the condition y>0, we finally get
y4+6110.

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