Previous Year Questions

Equation 2:

by² - 8y + 4 = 0

One of the roots of equation 2 is 2/3. So,

b(2/3)² - 8(2/3) + 4 = 0

4b/9 = 4/3

b = 3

Now, 3y² - 8y + 4 = 0

3y² - 6y - 2y + 4 = 0

3y(y - 2) - 2(y - 2) = 0

(y - 2)(3y - 2) = 0

y = 2, 2/3

The ratio of highest root of equation 1 to the highest root of equation 2 is 7: 4. So,

Highest root of equation 1 = 2 * 7/4 = 7/2

Now, Equation 1: ax² - 9x + 7 = 0

a(7/2)² - 9(7/2) + 7 = 0

49a/4 = 49/2

a = 2

So, 2x² - 9x + 7 = 0

2x² - 7x - 2x + 7 = 0

x(2x - 7) - 1(2x - 7) = 0

(2x - 7)(x - 1) = 0

x = 7/2, 1

(a + b)²x² + (3ab + 1)x - 6 = 0

(2 + 3)²x² + (3 * 2 * 3 + 1)x - 6 = 0

25x² + 19x - 6 = 0

25x² + 25x - 6x - 6 = 0

25x(x + 1) - 6(x + 1) = 0

(x + 1)(25x - 6) = 0

x = -1, 6/25