Given the two quadratic equations:

(i) $21x^2 + 10x + 1 = 0$

(ii) $24y^2 + 26y + 5 = 0$

Find the correct relationship between the roots $x$ and $y$ of these equations.

Question

Given the two quadratic equations:

(i) $21x^2 + 10x + 1 = 0$

(ii) $24y^2 + 26y + 5 = 0$

Find the correct relationship between the roots $x$ and $y$ of these equations.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$21x^2 + 10x + 1 = 0$

Try factoring:

$21x^2 + 7x + 3x + 1 = 0$

Group terms:

$7x(3x + 1) + 1(3x + 1) = 0$

Factor out common term:

$(7x + 1)(3x + 1) = 0$

So, the roots are:

$$x = -\frac{1}{7}, \quad x = -\frac{1}{3}$$

2. Solve the second quadratic equation:

$24y^2 + 26y + 5 = 0$

Rewrite middle term:

$24y^2 + 20y + 6y + 5 = 0$

Group terms:

$4y(6y + 5) + 1(6y + 5) = 0$

Factor common term:

$(4y + 1)(6y + 5) = 0$

Thus, the roots are:

$$y = -\frac{1}{4}, \quad y = -\frac{5}{6}$$

3. Compare the roots:

- Roots of $x$: $-\frac{1}{7} \approx -0.1429$, $-\frac{1}{3} \approx -0.3333$
- Roots of $y$: $-\frac{1}{4} = -0.25$, $-\frac{5}{6} \approx -0.8333$

There is no consistent inequality between $x$ and $y$ because:

- $-\frac{1}{7} > -\frac{1}{4}$
- $-\frac{1}{3} < -\frac{1}{4}$

Hence, no definite relation can be established between $x$ and $y$.

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

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