Question

If (a²+b²)x²+2(ab+bd)x+c²+d²=0 has no real roots, then
(a)ab=bc
(b)ab=cd
(c)ac=bd
(d)ad≠bc

Answer 1

SOLUTION :  

Option (d) is correct :  ad ≠ bc  

Given : (a² + b²)x² - 2(ac + bd)x + ( c² + d²) = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = (a² + b²) , b = - 2( ac + bd)  , c = ( c² + d²)

D(discriminant) = b² – 4ac

= {- 2(ac + bd)}² - 4(a² +b²)(c² + d²)  

= 4(ac + bd)² - 4(a² + b²)(c²+ d²)  

= 4(a²c²+ b²d² + 2abcd ) - 4( a²c² + a²d² + b²d² +  b²c²  

[(a + b)² = a² + b² + 2ab]

= 4(a²c² + b²d² + 2abcd  - a²c² -  a²d² - b²d² -  b²c² )  

= (a²c² - a²c² + b²d² - b²d² + 2abcd  -  a²d² -  b²c² )  

= 2abcd  -  a²d² -  b²c²  

= -(a²d² + b²c² - 2abcd)  

= a²d² + b²c² - 2abcd  

= (ad)² + (bc)² - 2×ad × bc  

= (ad - bc)²  

[(a - b)² = a² + b² - 2ab]

Given roots are not real so, D < 0

(ad - bc)² <  0

(ad - bc) <  0

ad ≠ bc  

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Concept used:

We know, for no real roots,

Calculation:

Given equation is (a2 + b2) x2 + 2(ab + bd) x + c2 + d2 = 0

Here, a = (a2 + b2), b = 2(ab + bd), c = c2 + d2

Therefore,

(ad - bc) < 0

or ad < bc

∴ ad ≠ bc