If two roots of the equation, $x^{3}-7x^{2}+4x+12=0$ are in the ratio $1:3 $, then the roots are
- $-6,-2,-1$
- $-1,2,6$
- $1,3,4 $
- $-2,3,6$
Let the roots be a, 3a, and b. By Vieta's formulas: a + 3a + b = 7 => 4a + b = 7. a * 3a * b = -12 => 3a^2 * b = -12 => a^2 * b = -4. Substituting b = 7 - 4a into the second: a^2(7 - 4a) = -4 => 4a^3 - 7a^2 - 4 = 0. Testing a = -1 gives 4(-1) - 7(1) - 4 = -15 (no). Testing a = 2 gives 4(8) - 7(4) - 4 = 32 - 28 - 4 = 0. So a=2, 3a=6, b=7-8=-1. The roots are -1, 2, 6.
Let the three roots of the cubic equation be a, 3a, and b. Using Vieta's formulas for the equation x cubed minus 7x squared plus 4x plus 12 equals 0, we know the sum of the roots is 7, so 4a plus b equals 7. We also know the product of the roots is negative 12, so 3a squared times b equals negative 12. Solving the system by substituting b equals 7 minus 4a into the product equation yields 3a squared times (7 minus 4a) equals negative 12, which gives the valid roots negative 1, 2, and 6.