Question

If ad is not equal to bc, then prove that the equation (a2 +b2) x2+2 (ac +bd) x + (c2 d2)- 0 has no real roots.

Answer 1

Compare the given quadratic equation with Ax2 + Bx + C = 0
Here A = a2 + b2 , B = 2(ac + bd) and C = c2 + d2 
Consider, B2 – 4AC = [2(ac + bd)]2 – 4 ´ (a2 + b2 ) x (c2 +d2) 
= 4[a2c2 + 2abcd + b2d2] – 4[a2c2 + a2d2 + b2c2 + b2d2]
= 4a2c2 + 8abcd + 4b2d2 – 4a2c2 – 4a2d2 – 4b2c2 – 4b2d2
= 8abcd– 4a2d2 – 4b2c2
= –4[4a2d2 + 4b2c2 – 2abcd]
= –4[ad – bc]2

Answer 2

Answer:

Step-by-step explanation:

Solution :-

We have, A = (a² + b²), B = 2(ac + bd) and C = (c² + d²)

For no real roots, D < 0

i.e, D ⇒ b² - 4ac = 0

b² - 4ac

= [2(ac + bd)]² - 4(a² + b²)(c² + d²)

= 4[a²c² + 2abcd + b²d²] - 4[a²c² + a²d² + b²c² + b²d²]

= 4[a²c² + 2abcd + b²d² - a²c² - a²d² - b²c² - b²d²]

= - 4[a²d² + b²c² - 2abcd]

= - 4[ad - bc]²

Since, ad ≠ bc

Therefore, D < 0

Hence, the equation has no real roots.