3x^2 + 64 = 2x^2/2 find the roots of the quadratic equation
1 answer:
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Answer:
y = (x/(1-x))√(1-x²)
Step-by-step explanation:
The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:
x = r·cos(θ)
y = r·sin(θ)
x² +y² = r²
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r = sec(θ) -2cos(θ)
r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)
r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²
(x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations
x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor
y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)
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The attached graph shows the equivalence of the polar and rectangular forms.
