Given the two quadratic equations:

(i) $12x^2 + 11x + 2 = 0$

(ii) $12y^2 + 7y + 1 = 0$

Find the correct relationship between the roots $x$ and $y$.

Question

Given the two quadratic equations:

(i) $12x^2 + 11x + 2 = 0$

(ii) $12y^2 + 7y + 1 = 0$

Find the correct relationship between the roots $x$ and $y$.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$12x^2 + 11x + 2 = 0$$
   Factorizing:
   $$12x^2 + 8x + 3x + 2 = 0$$
   $$4x(3x + 2) + 1(3x + 2) = 0$$
   $$(4x + 1)(3x + 2) = 0$$
   So, the roots are:
   $$x = -\frac{1}{4}, \quad x = -\frac{2}{3}$$

2. Solve the second quadratic equation:
   $$12y^2 + 7y + 1 = 0$$
   Factorizing:
   $$12y^2 + 4y + 3y + 1 = 0$$
   $$4y(3y + 1) + 1(3y + 1) = 0$$
   $$(4y + 1)(3y + 1) = 0$$
   So, the roots are:
   $$y = -\frac{1}{4}, \quad y = -\frac{1}{3}$$

3. Comparing roots:
   - Roots of $x$ are $-\frac{1}{4}$ and $-\frac{2}{3}$
   - Roots of $y$ are $-\frac{1}{4}$ and $-\frac{1}{3}$

4. Since one root $-\frac{1}{4}$ is common, but the other roots $-\frac{2}{3}$ and $-\frac{1}{3}$ differ, no definite relation such as $x > y$ or $x < y$ can be established between $x$ and $y$.

**Answer:** (b) $x = y$ or no relation can be established between $x$ & $y$.

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