Question

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

Answer 1

please see answer in the pic given below Mark it as a brainliest answer

Answer 2

Gɪᴠᴇɴ

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

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Tᴏ ᴘʀᴏᴠᴇ

➳ ac + bd = 0

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Sᴛᴇᴘꜱ

❍ 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

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

Compare the above equation with

☞ a² + b² + 2ab = (a+b)²

We get ;

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

◕ Hence, it is proved.

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Identities used :

(a - b)² = a² - 2ab + b²

(a + b)² = a² + 2ab + b²

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