The value of $a$ for which one root of the quadratic equation $(a^2-5a+3) x^2+(3a-1)x+2=0 $ is twice as large as the other, is :
- $-\cfrac{2}{3}$
- $\cfrac{1}{3}$
- $-\cfrac{1}{3}$
- $\cfrac{2}{3}$
Let roots be alpha and 2*alpha. Sum of roots = -b/a = -(3a-1)/(a^2-5a+3) = 3*alpha. Product of roots = c/a = 2/(a^2-5a+3) = 2*alpha^2. From product, alpha^2 = 1/(a^2-5a+3). From sum, alpha = -(3a-1)/(3*(a^2-5a+3)). Squaring alpha: (3a-1)^2 / (9*(a^2-5a+3)^2) = 1/(a^2-5a+3). (9a^2 - 6a + 1) = 9*(a^2-5a+3) = 9a^2 - 45a + 27. 39a = 26. a = 26/39 = 2/3.
Let the roots of the quadratic equation (a^2-5a+3)x^2 + (3a-1)x + 2 = 0 be r and 2r. Using the sum of roots formula, r + 2r = -(3a-1) / (a^2-5a+3), which simplifies to 3r = (1-3a) / (a^2-5a+3). Using the product of roots formula, r * 2r = 2 / (a^2-5a+3), which simplifies to 2r^2 = 2 / (a^2-5a+3), meaning r^2 = 1 / (a^2-5a+3). From the sum equation we isolate r as r = (1-3a) / [3(a^2-5a+3)], and substituting this into the squared equation gives (1-3a)^2 / [9(a^2-5a+3)^2] = 1 / (a^2-5a+3). Multiplying both sides by 9(a^2-5a+3)^2 yields (1-3a)^2 = 9(a^2-5a+3), and expanding the binomial gives 1 - 6a + 9a^2 = 9a^2 - 45a + 27. Canceling 9a^2 from both sides leaves 1 - 6a = -45a + 27, and moving terms gives 39a = 26, which results in a = 2/3.