Question

show that the roots of the quadratic equationsare equal then ad = bc. X2(A2 + b2) +2(ac + bd)x +( c2+ D2)​

Answer 1

Step-by-step explanation:

$\bf \large \green{given}$

$\bf \large \{{x}^{2}( {a}^{2} + {b}^{2} ) +2(ac + bd)x +( {c}^{2} + {d}^{2} )$

$\bf \large \red{to \: prove}$

$\bf \: ad = bc$

$\bf \large \orange{solution}$

equation has real roots .

Therefore, discriminant =0

$\large\bf \: d \: = {b}^{2} + 4ac$

Thus, [2(ac + bd)]² + 4(a² + b²) × (c² + d²)

⇒ 4(a²c² + b²d² +2ac.bd)= 4(a²c² + a²d² + b²c² + b²d²)

⇒ (ad - bc)² = 0

⇒ad−bc=0

⇒ad=bc

Answer 2

Answer:

Quadratic equations are in the form of,

x²-(α+β)x+αβ = 0

f(x) = (a²+b²)x² + 2(ac+bd)x+(c²+d²) = 0

α+β = -2(ac+bd)/(a²+b²)

αβ = (c²+d²)/(a²+b²)

Roots of the quadratic equations are equal, α = β (given)

α+β = -2(ac+bd) / (a²+b²)

2α = -2(ac+bd) / (a²+b²)

α = (-ac-bd) / (a²+b²)

αβ = (c²+d²) / (a²+b²)

α.α = (c²+d²) / (a²+b²)

α² = (c²+d²) / (a²+b²)

(-ac-bd)²/(a²+b²) = (c²+d²) / (a²+b²)

(ac+bd)²/(a²+b²)² = (c²+d²) / (a²+b²)

(ac+bd)²/(a²+b²) = (c²+d²)

(ac+bd)² = (a²+b²)(c²+d²)

(ac)²+(bd)²+2abcd = (a²+b²)(c²+d²)

(ac)²+(bd)²+2abcd = (ac)²+(ad)²+(bc)²+(bd)²

2abcd = (ad)²+(bc)²

(ad)²+(bc)²-2abcd = 0

(ac-bd)² = 0

ac-bd = 0

ac = bd

Hence it proved .