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in Quadratic Equations by (49.4k points)
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Find the real values of x and y for which:

(1 + i) y2 + (6 + i) = (2 + i)x

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Given: (1 + i) y2 + (6 + i) = (2 + i)x

Consider, (1 + i) y2 + (6 + i) = (2 + i)x

⇒ y2 + iy2 + 6 + i = 2x + ix

⇒ (y2 + 6) + i(y2 + 1) = 2x + ix

Comparing the real parts, we get

y2 + 6 = 2x

⇒ 2x – y2 – 6 = 0 …(i)

Comparing the imaginary parts, we get

y2 + 1 = x

⇒ x – y2 – 1 = 0 …(ii)

Subtracting the eq. (ii) from (i), we get

2x – y2 – 6 – (x – y2 – 1) = 0

⇒ 2x – y2 – 6 – x + y2 + 1 = 0

⇒ x – 5 = 0

⇒ x = 5

Putting the value of x = 5 in eq. (i), we get

2(5) – y2 – 6 = 0

⇒ 10 – y2 – 6 = 0

⇒ -y2 + 4 = 0

⇒ - y2 = -4

⇒ y2 = 4

⇒ y = √4

⇒ y = ± 2

Hence, the value of x = 5 and y = ± 2

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