Multiple choice

If $a,b,c$ are postive rational numbers such that $a > b > c$ and the quadratic equation $(a+b-2c)x^{2}+(b+c -2a)x+(c+a-2b)=0$ has a root in the interval $(-1,0)$, then

  1. $b-c > a$
  2. $c+a <2b$

  3. both roots of given equation are rational
  4. the equation $ax^{2}+2bx+c=0$ has both negative real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) be the quadratic. Its leading coefficient is positive, and f(-1) = 2(2a - b - c) > 0 because a > b > c. A root in (-1,0) therefore requires f(0) = c + a - 2b < 0, so c + a < 2b.

AI explanation

Let f(x) be the given quadratic expression. We evaluate f(-1) and f(0) to find f(-1) = (a+b-2c)(1) + (b+c-2a)(-1) + (c+a-2b), which simplifies to 4(b-c). We also find f(0) = c+a-2b. Because f(x) is a continuous function with a root between -1 and 0, the values f(-1) and f(0) must have opposite signs, making their product negative. Substituting our evaluated expressions gives 4(b-c)(c+a-2b) < 0. Since the problem states a > b > c, the difference b-c is strictly positive. Dividing the inequality by the positive quantity 4(b-c) forces the second term to be negative, yielding c+a-2b < 0. Adding 2b to both sides results in c+a < 2b.

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