Question

If the roots of the quadratic equations (a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0 are equal. Prove that a/b=c/d

Answer 1

Question

If the roots of the quadratic equations (a² + b²)x² -2(ac + bd)x + (c² + d²) = 0 are equal. Prove that a/b = c/d

To Prove

a/b = c/d

Proof

Given quadratic equation is, (a² + b²)x² -2(ac + bd)x + (c² + d²) = 0

The roots of the above quadratic equation are equal.

If roots are equal then, D = b² - 4ac = 0

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

Substitute these values in the above formula

→ [-2(ac + bd)]² - 4[(a² + b²)(c² + d²)] = 0

→ 4(a²c² + b²d² + 2ac.bd) - 4(a²c² + a²d² + b²c² + b²d²) = 0

Used identity: (a + b)² = a² + b² + 2ab

→ 4(a²c² + b²d² + 2ac.bd) - 4(a²c² + a²d² + b²c² + b²d²) = 0

Take 4 as common

→ 4[(a²c² + b²d² + 2ac.bd) - (a²c² + a²d² + b²c² + b²d²)] = 0

→ (a²c² + b²d² + 2ac.bd) - (a²c² + a²d² + b²c² + b²d²) = 0

→ a²c² + b²d² + 2ac.bd - a²c² - a²d² - b²c² - b²d² = 0

→ a²c² - a²c² + b²d² - b²d² + 2ac.bd - a²d² - b²c² = 0

a²c² - a²c² + b²d² - b²d², will cancel out

We left with

→ 2ac.bd - a²d² - b²c² = 0

Take ' - ' as common

→ -(-2ac.bd + a²d² + b²c²) = 0

→ - 2ac.bd + a²d² + b²c² = 0

→ (ad - bc)² = 0

Used identity: (a - b)² = a² + b² - 2ab

→ ad - bc = 0

→ ad = bc

→ a/b = c/d

Hence, proved

Answer 2

||✪✪ QUESTION ✪✪||

If the roots of the quadratic equations (a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0 are equal. Prove that a/b=c/d ?

|| ★★ CONCEPT USED ★★ ||

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

|| ✰✰ ANSWER ✰✰ ||

Given Equation is :- (a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0

Here , we have ,

⟿ A = (a² + b²)

⟿ B = -(2)[ac+bd]

⟿ C = (c² + d²)

Since, Roots are Equal , D = 0 ,

➪ B² - 4AC = 0

Putting values we get :-

➼ [-2(ac+bd)]² - 4 * (a²+b²) * (c²+d²) = 0

➼ 4 [ac + bd)²] - 4 * (a²+b²) * (c²+d²) = 0

➼ 4[a²c² + b²d² + 2abcd ] - 4 [ a²c² + a²d² + b²c² + b²d²] = 0

➼ 4 [a²c² -a²c² + b²d² - b²d² + 2abcd -a²d² - b²c²] = 0

➼ 4 [ 2abcd -a²d² - b²c²] = 0

Taking 4 in RHS side, and Taking (-1) Common From LHS,

➼ a²d² + b²c² - 2abcd = 0

Comparing it with [a² + b² - 2ab = (a-b)² ] , we get,

➼ (ad - bc)² = 0

➼ ad - bc = 0

➼ ad = bc

➼ (a/b) = (c/d)

☘☘☘ Hence, Proved. ☘☘☘