Math Olympiad Test: Quadratic Equations- 3 - Class 10 MCQ

 ...(i)
Squaring both sides of (1), we get

Since x cannot be negative, therefore, neglect 
Thus, x = 2

Given equation is ax² + bx + c = 0
Roots are real and unequal, if b² – 4ac > 0

Given equation is a(b – c)x² + b(c – a)x + c(a – b) = 0
Let α be the other root, then
Product of roots = 

Since x = 2 is a root of the equation
x² + bx + 12 = 0 ⇒ (2)² + b(2) + 12 = 0
⇒ 2b = –16 ⇒ b = –8
Then, the equation x² + bx + q becomes
x² – 8x + q = 0 ...(1)
Since (1) has equal roots ⇒ b² – 4ac = 0
⇒ (–8)² – 4(1)q = 0 ⇒ q = 16

Let the equation be x² + ax + b = 0
Its roots are 3 and 2
∴ Sum of roots, 5 = –a
and product of roots, 6 = b
∴ Equation is x² – 5x + 6 = 0
Now constant term is wrong and it is given that correct constant term is – 6.
∴ x² – 5x – 6 = 0 is the correct equation.
Its roots are –1 and 6.

Since the given equation has equal roots.
∴ D = 0
⇒ (b – c)² – 4(c – a) (a – b) = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (–2a)² + (b)² + (c)² + 2(–2a)(b) + 2bc + 2(–2a)(c) = 0
⇒ (–2a + b + c)² = 0
⇒ –2a + b + c = 0 ⇒ 2a = b + c

Equation is 2x² + ax + 32 = 0
Let one root be α, then other would be 2α.
Now, α × 2α = 16 ⇒ α = ±2√2
and α + 2α = –a/2 ⇒ 3α = –a/2 ⇒ 6α = –a
⇒ ±12√2 = −a or a = ±12√2

Given equation is  ...(1)
Rationalising (1), we get 
 ...(2)
Adding (1) and (2), we get

Given equation is x² + x – (a + 1)(a + 2) = 0
⇒ x² + (a + 2)x – (a + 1)x – (a + 1) (a + 2) = 0
⇒ x(x + (a + 2)) – (a + 1)(x + (a + 2)) = 0
⇒ (x – (a + 1))(x + (a + 2)) = 0
⇒ x = (a + 1) or x = – (a + 2)

Given equation is 2x² + 6x + a = 0
Now, 
⇒ 
⇒  
⇒ (9/2) < a