Question

if the roots of quadratic equation (a^+b^)x^+2( bc -ad)+c^+d^=0 are equal .show that ac+ bd=0​

Answer 1

Given equation :

• (a² + b²)x² + 2(bc - ad)x + (c² + d²)

To prove :

• ac + bd = 0

Proof :

We know that,

The standard form of the quadratic equation is ax² + bx + c = 0. Here,

• a = (a² + b²)
• b = 2(bc - ad)
• c = (c² + d²)

It is given that, the roots of the equation are zero. Thus, discriminant is zero.

Discriminant = b² - 4ac = 0

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

Compare the above equation with

• a² + b² + 2ab = (a+b)²

We get ;

$\Rightarrow \sf (ac+bd)^2 = 0\\\\\Rightarrow \sf ac + bd = 0$

Hence, it is proved.

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Identities used :

• (a - b)² = a² - 2ab + b²
• (a + b)² = a² + 2ab + b²

Answer 2

Concept :-

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.

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Solution :-

Since Roots Are Equal . So, D = 0 .

→ (a² + b²) x² + 2(bc – ad) x + (c² + d²) = 0 = A•x^2 + B•x + C = 0

comparing we get :-

→ A = (a² + b²)

→ B = 2(bc – ad)

→ C = (c² + d²)

Putting all values in D = 0 Now, we get ,

→ B^2 - 4•A•C = 0

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

→ [4(b²c² + a²d² - 2abcd)] - 4[a²c² + a²d²+ b²c² + b²d²] = 0

→ [4b²c² + 4a²d² - 8abcd] - [4a²c² + 4a²d²+ 4b²c² + 4b²d²] = 0

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

→ - 8abcd - 4a²c² - 4b²d² = 0

→ -4(2abcd + a²c² + b²d²) = 0

→ a²c² + b²d² + 2abcd = 0

comparing with a² + b² + 2ab = (a + b)² , we get,

→ [ ac + bd]² = 0

Hence,

→ (ac + bd) = 0 (Proved).