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Question 12:Determine whether the given quadratic equation has root(s). If so, firroot(s): 3y2 + (6 + 4a)y + 8a = 0. |
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Answer» 3y^2+(6+4a)y+8a; a=3; b=6+4a, c=8a, y=-b+-Vb^2-4ac/2a; y=-(6+4a)+-V(6+4a)^2-4(3)(8a)/2(3)= -6-4a+V36+48+a^2-96a/6= -6-4a+-V36+48a+a^2-96/6= y=-6-4a+-V6^2-48a+36/6; y=-6+4a+22a-36; y=-6+4a-22a-36=126, y=-2 (or)-4a^3 |
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