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# If the height of an equilateral triangle is also a root pf the equation y^4-3y^2-270, then the area of the triangle is?

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- 2 years agoFavorite Answer
y^4 - 3y^2 - 270 = 0

y^2 = (3 +/- sqrt(9 + 1080)) / 2

y^2 = (3 +/- sqrt(1089)) / 2

y^2 = (3 +/- 33) / 2

y^2 > 0

y^2 = (3 + 33) / 2

y^2 = 36/2

y^2 = 18

y = +/- 3 * sqrt(2)

y = 3 * sqrt(2)

So, the height is equal to 3 * sqrt(2)

(s/2)^2 + h^2 = s^2

(s/2)^2 + 18 = s^2

s^2 / 4 + 18 = s^2

18 = (3/4) * s^2

18 * (4/3) = s^2

24 = s^2

2 * sqrt(6) = s

A = (1/2) * s * h

A = (1/2) * 2 * sqrt(6) * sqrt(18)

A = sqrt(6) * sqrt(6) * sqrt(3)

A = 6 * sqrt(3)

- PinkgreenLv 72 years ago
y^4-3y^2-270=0

=>

y^2=18 (take +ve root only)

=>

y=3sqr(2) (take +ve root only)

Let a be the side-length of the triangle, then

asin(60*)=3sqr(2)

=>

a=2sqr(6)

=>

the area of the triangle=

2sqr(6)*3sqr(2)/2=6sqr(3)

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