Multiple choice

If the roots of the equation $x^2-2cx+ab=0$ be real and unequal, the roots of the equation ${x^2} - 2(a + b)x + {a^2} + {b^2} + 2{c^2} = 0$ are

  1. real and distinct

  2. real and equal

  3. real

  4. imaginary
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Roots of x^2 - 2cx + ab = 0 are real/unequal, so discriminant (2c)^2 - 4ab > 0 => c^2 > ab. For the second equation, discriminant D = [2(a+b)]^2 - 4(a^2 + b^2 + 2c^2) = 4(a^2 + 2ab + b^2) - 4a^2 - 4b^2 - 8c^2 = 8ab - 8c^2 = 8(ab - c^2). Since c^2 > ab, ab - c^2 < 0, so D < 0. Roots are imaginary.

AI explanation

For the first equation x^2 - 2cx + ab = 0, the condition for real and unequal roots requires its discriminant to be positive, meaning (-2c)^2 - 4ab > 0, so c^2 > ab. The discriminant of the second equation x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0 is calculated as D = [-2(a + b)]^2 - 4(1)(a^2 + b^2 + 2c^2). Simplifying this gives D = 4(a^2 + 2ab + b^2 - a^2 - b^2 - 2c^2) = 8ab - 8c^2 = 8(ab - c^2). Because c^2 is strictly greater than ab, the value of (ab - c^2) is negative, making the discriminant negative and proving the roots are imaginary.

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