Multiple choice

If the quadratic equation $a{ x }^{ 2 } + bx + c = 0; a>0$ has real roots of opposite sign in the interval $\left( -2, 2 \right)$, then comment on the value of the following expression $1 + \displaystyle\frac { c }{ 4a } - \left| \displaystyle\frac { b }{ 2a } \right| $.

  1. $(0,\infty)$
  2. $(-\infty,0)$
  3. $(1-\sqrt{3},1+\sqrt{3})$
  4. $(-\infty,1-\sqrt{3})\cup (1+\sqrt{3},\infty)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given roots are real and of opposite sign, c/a < 0. The expression is 1 + c/4a - |b/2a|. Since the roots are in (-2, 2), the sum of roots -b/a is in (-4, 4), so |b/2a| < 2. Also, c/a is the product of roots, so c/a > -4. The expression simplifies to values in (0, infinity).

AI explanation

Since the quadratic equation has real roots of opposite sign, the sum of the roots lies strictly between -2 and 2, giving the absolute value of b/(2a) as strictly less than 1. Both roots lie in the interval (-2, 2), so the polynomial evaluated at 2 and -2 is strictly positive; this means 4a plus or minus 2b plus c is greater than 0. Dividing this inequality by 4a yields 1 plus or minus b/(2a) plus c/(4a) is greater than 0. Let k equal b/(2a); since the absolute value of k is strictly less than 1, 1 minus the absolute value of k is strictly positive. Because 1 plus c/(4a) is strictly greater than the absolute value of k, subtracting the absolute value of k from both sides proves that 1 plus c/(4a) minus the absolute value of b/(2a) is greater than 0. The result is (0, infinity).

Keep practicing — related questions