Let , be the roots of the equation x^2 - 3x + r = 0 , and 2 , 2 be the roots of the equation x^2 + 3x + r = 0 . If the roots of the equation x^2 + 6x = m are 2 + + 2r and - 2 - r 2 , then m is equal to:

Step-by-step solution

From the first equation x^2 - 3x + r = 0 , the sum and product of the roots are: + = 3 = r From the second equation x^2 + 3x + r = 0 , the sum and product of the roots are: 2 + 2 = -3 ( 2 )(2 ) = r = r We have a system of linear equations in terms of and : + = 3 + 4 = -6 Subtracting the first equation from the second gives: 3 = -9 = -3 Substituting = -3 into + = 3 gives: = 6 Now, we can find r : r = = 6 (-3) = -18 The roots of the third equation x^2 + 6x - m = 0 are given as 2 + + 2r and - 2 - r 2 . Let us calculat