Question

If the roots of the quadratic equation (a²+b²)x² +2(bc-ad)x +c²+d² =0 are real and equal then show that ac + bd = 0 .​

Answer 1

Step-by-step explanation:

Given:-

• A quadratic equation (a² + b²)x² + 2(bc - ad)x + c² + d² = 0.

• The roots of the equation are real and equal.

To Prove:-

• ac + bd = 0

Concept used:-

For a quadratic equation ax² + bx + c = 0

If the roots are real and equal

Then, b² - 4ac = 0

Solution:-

Comparing (a² + b²)x² + 2(bc - ad)x + c² + d² =0 with ax² + bx + c

Here:-

• a = a² + b²

• b = 2(bc - ad)

• c = c² + d²

$\sf \implies {b}^{2} - 4ac = 0$

$\sf \implies { \big[2(bc - ad) \big]}^{2} - 4( {c}^{2} + {d}^{2})( {a}^{2} + {b}^{2} ) = 0$

$\sf \implies { 4(bc - ad)}^{2} = 4( {c}^{2} + {d}^{2})( {a}^{2} + {b}^{2} )$

$\sf \implies {(bc - ad)}^{2} = ( {c}^{2} + {d}^{2})( {a}^{2} + {b}^{2} )$

$\sf \implies {b}^{2} {c}^{2} + {a}^{2} {d}^{2} + 2acbd = {a}^{2} {c}^{2} + {a}^{2} {d}^{2} + {b}^{2} {c}^{2} + {b}^{2} {d}^{2}$

$\sf \implies {b}^{2} {c}^{2} + {a}^{2} {d}^{2} - {a}^{2} {d}^{2} - {b}^{2} {c}^{2} = {a}^{2} {c}^{2} + {b}^{2} {d}^{2} + 2acbd$

$\sf \implies {a}^{2} {c}^{2} + {b}^{2} {d}^{2} + 2acbd = 0$

$\sf \implies (ac + bd)^{2} = 0$

$\sf \implies ac + bd= 0$

$\dashrightarrow \large\underline{\boxed{\therefore \textsf{\textbf{ac + bd = 0}}}}$

Hence Proved

Answer 2

$\huge\sf\underline\red{SOLUTION}$

Given that,

Roots of quadratic equation are real and equal

Roots of (a² +b²) x² + 2(bc - ad ) x + c²+ d²=0

Show that ac + bd =0

So, If roots are real and equal b²-4ac =0

So,

a = a² + b²

b = 2(bc - ad)

c = c²+d²

So

b² -4ac =0

Plugging values of a,b,c

[2(bc-ad)]² - 4 ( a² + b² ) (c² + d²)=0

4 ( bc - ad )² - 4 ( a² + b² ) c² + (a² + b²)(d²)=0

4( b²c²-2bcad + a²d²) -4(a²c² + b²c²)+(a²d²+b²d²)=0

4b²c²-8abcd+4a²d²-4a²c²-4b²c²-4a²d²-4b²d²=0

-4a²b²d² -4a²c² -8abcd

-4(2abcd + a²c² +b²d²)=0

2abcd + a²c²+b²d²=0

(ac + bd)² =0

ac + bd =0. Hence proved!!!!

KNOW MORE

If D=0 roots are real and equal

If D >0 roots are real and distinct

If D<0 roots are complex and conjugate