math help.... assignment..?

Question

can you please find the roots of this equation.. thanks.. pease include the step by step solution....

180^2 (y^2 + 4y + 69)^3 = 100^2 (y^2 - 8y + 29)^3

here's how it reads....

one hundred eighty two squared times open quantity of y squared plus four y plus sixty nineclose quantity cube equals one hundred squared open quantity y squared minus eight y plus 29 close quantity cube.. the expression y^2+4y+69 and y^2 -8y +29 are the only ones cubed.. the 180 and 100 are squared... got it????? thanks

Answer 1

180^2 (y^2 + 4y + 69)^3 = 100^2 (y^2 - 8y + 29)^3
y^2 + 4y + 69 = 0.6758(y^2 - 8y + 29)
y^2 + 4y + 69 = 0.6758y^2 - 8*0.6758y + 29*0.6758
0.3242y^2 + 9.4064y + 63.594 = 0
y^2 + 29.014y + 196.16 = 0
y ≈ (- 29.014 ± √(29.014^2 - 4*196.16))/2
y ≈ (- 29.014 ± √57.172196)/2
y ≈ (- 29.014 ± 7.561)/2
y ≈ - 18.288, -10.726

Answer 2

I dont really understand your qn.

is it (180^2(y^2 + 4y + 69))^3 = (100^2 (y^2 - 8y + 29))^3 or is it 180^(2 (y^2 + 4y + 69)^3) = 100^(2 (y^2 - 8y + 29)^3). both of them are different, u know...but lets try both...the first one first.

180^2 (y^2 + 4y + 69)^3 = 100^2 (y^2 - 8y + 29)^3
180^2 (y^2 + 4y + 69) = 100^2 (y^2 - 8y + 29)
180^ y^2 + 4y + 69 = 100^ y^2 - 8y + 29

hey...wait a minute, did you copy the qn wrongly...it does not work out....