Multiple choice

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

For the first equation, the discriminant D1 = (-2c)^2 - 4(ab) = 4(c^2 - ab) > 0, so c^2 > ab. For the second equation, the discriminant D2 = [-2(a+b)]^2 - 4(a^2 + b^2 + 2c^2) = 4(a^2 + 2ab + b^2 - a^2 - b^2 - 2c^2) = 4(2ab - 2c^2) = 8(ab - c^2). Since c^2 > ab, ab - c^2 < 0, meaning D2 < 0. Therefore, the roots are imaginary.

AI explanation

The discriminant of the second equation, x^2 - 2(a+b)x + (a^2 + b^2 + 2c^2) = 0, is found using D = [-2(a+b)]^2 - 4(1)(a^2 + b^2 + 2c^2). Simplifying this gives D = 4(a^2 + 2ab + b^2) - 4a^2 - 4b^2 - 8c^2, which reduces to 8ab - 8c^2 or 8(ab - c^2). Because the first equation has real and unequal roots, its discriminant is positive, meaning 4c^2 - 4ab > 0 and c^2 > ab. This makes the discriminant of the second equation negative (8 times a negative value), which means its roots are imaginary.

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