Test: Quadratic Equations (April 29) - CAT MCQ

For the given expression to be a maximum, the denominator should be minimized. (Since, the function in the denominator has imaginary roots and is always positive), x2 + 5x + 10 will be minimized at x = -2.5 and its minimum values at x = -2.5 is 3.75.
Hence, required answer = 1/3.75 =4/15.

GIVEN:

 f(x) = x² - 14x + 45

FORMULA USED:

Sum of roots = (-b/a) and Product of roots = c/a for f(x) = ax² + bx + c 

CALCULATION:

f(x) = x² - 14x + 45

⇒ a = 1, b = -14, c = 45

⇒ Sum of roots(P + Q) = (-b/a)

⇒ 14

⇒ Product of roots (PQ) = c/a

⇒ 45

⇒ (1/P + 1/Q) = (P + Q)/PQ

⇒ 14/45

∴  (1/P + 1/Q) = 14/45

For a, b negative the given expression will always be positive since, a², b² and ab are all positive.

x² + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
=> x = 3 or x = -5.

Solve this by assuming each option to be true and then check whether the given expression has equal roots for the option under check.

Thus, if we check for option (b). ad = be.
We assume a = 6, d = 4 b = 12 c = 2 (any set o f values that satisfies ad = bc).
Then (a² + b²) x² - 2(ac + bc) x + (c² + d²) = 0

180 x²- 120 x +20 = 0.
We can see that this has equal roots. Thus, option (b) is a possible answer. The same way if we check for a, e and d we see that none of them gives us equal roots and can be rejected.

Solving quadratically, we have option (b) as the root of this equation.

Solve by assuming values of a, b, and c in AP, GP and HP to check which satisfies the condition.