Given the two quadratic equations:

(i) $ 2x^2 + 9x + 9 = 0 $

(ii) $ 15y^2 + 16y + 4 = 0 $

Find the correct relationship between $ x $ and $ y $ based on their roots.

Question

Given the two quadratic equations:

(i) $ 2x^2 + 9x + 9 = 0 $

(ii) $ 15y^2 + 16y + 4 = 0 $

Find the correct relationship between $ x $ and $ y $ based on their roots.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$ 2x^2 + 9x + 9 = 0 $$

Factorize by grouping:

$$ 2x^2 + 6x + 3x + 9 = 0 $$

$$ 2x(x + 3) + 3(x + 3) = 0 $$

$$ (2x + 3)(x + 3) = 0 $$

So, the roots are:

$$ x = -\frac{3}{2} \quad \text{or} \quad x = -3 $$

2. Solve the second quadratic equation:

$$ 15y^2 + 16y + 4 = 0 $$

Factorize by grouping:

$$ 15y^2 + 10y + 6y + 4 = 0 $$

$$ 5y(3y + 2) + 2(3y + 2) = 0 $$

$$ (5y + 2)(3y + 2) = 0 $$

So, the roots are:

$$ y = -\frac{2}{5} \quad \text{or} \quad y = -\frac{2}{3} $$

3. Compare the roots:

- For $ x $: roots are $ -1.5 $ and $ -3 $

- For $ y $: roots are $ -0.4 $ and $ -0.666... $

The largest root of $ x $ is $ -1.5 $, and the smallest root of $ y $ is $ -0.666... $

Since $ -1.5 < -0.666... $, it follows that $ x < y $.

**Answer:** $ x < y $

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