Multiple choice

Consider the quadratic equation $(1+m)x^2-2(1+3m)x+(1+8m)=0$, (where $m \in R-\left {-1\right })$, t hen the number of real values of $m$ such that the given quadratic equation has roots in the ratio $2: 3$ are,

  1. $0$
  2. $2$
  3. $3$

  4. Infinitely many
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be 2k and 3k. Sum = 5k = 2(1+3m)/(1+m). Product = 6k^2 = (1+8m)/(1+m). From sum, k = 2(1+3m)/(5(1+m)). Substitute into product: 6 * [4(1+3m)^2 / 25(1+m)^2] = (1+8m)/(1+m). This simplifies to a quadratic in m, yielding two values.

AI explanation

Let the roots be 2a and 3a, so their sum gives 5a equals 2 times 1 plus 3m divided by 1 plus m, and their product gives 6a squared equals 1 plus 8m divided by 1 plus m. Solving the sum equation for m yields m equals 3a minus 1. Substituting this expression for m into the product equation and simplifying produces the quadratic 6a squared minus 5a minus 1 equals 0. Factoring this equation gives the solutions a equals 1 and a equals negative 1 over 6, which lead to exactly two real values for m. The number of real values is 2.

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