Question

If the roots of the equation x2 + 2cx + ab are real unequal, prove that the
equation x2-2(a+b ) x + a2+b2+2c2has no real roots.

Answer 1

you can put b^2-4ac>0 for first equation
(2c)^2-4ab>0
4c^2-4ab>0
c^2-ab>0
c^2>ab
and than check b^2-4ac for equation second
than you get your ans.
and than any problem comment me..

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here we have,

x² + 2cx + ab = 0 .... (i)

x² - 2(a + b)x + a² + b² + 2c² = 0 .....(ii)

Let D₁ and D₂ be the discriminant of equation (i) and (ii)

Then,

⇒ D₁ = (2c)² - 4 × 1 × ab

⇒ D₁ = 4c² - 4ab

⇒ D₁ = 4(c² - ab)

And,

⇒ D₂ = [- 2(a + b)]² - 4 × 1 × (a² + b² + 2c²)

⇒ D₂ = 4(a + b)² - 4(a² + b² + 2c²)

⇒ D₂ = 4[a² + b² + 2ab - a² - b² - 2c²]

⇒ D₂ = 4(2ab - 2c²)

⇒ D₂ = - 8(c² - ab)

It is given that the roots of a equation (i) are real and unequal.

Therefore,

D₁ > 0

⇒ 4(c² - ab)  > 0

⇒ c² - ab  > 0

⇒ - 8(c² - ab) < 0

⇒ D₂ < 0

Hence, the roots of equation (ii) are not equal.