Multiple choice

Find the value of $p$ for which given quadratic has real roots. $\displaystyle 5{ x }^{ 2 }+8x+p=0$

  1. $\displaystyle p\ge \dfrac{5}{16}$
  2. $\displaystyle p\le \dfrac{5}{16}$
  3. $\displaystyle p\le \dfrac{-5}{16}$
  4. $\displaystyle p\le \dfrac{16}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a quadratic equation ax^2 + bx + c = 0 to have real roots, the discriminant D = b^2 - 4ac must be >= 0. Here, 8^2 - 4(5)(p) >= 0, which means 64 - 20p >= 0. Thus, 64 >= 20p, or p <= 64/20 = 16/5.

AI explanation

To ensure the quadratic equation 5x^2 + 8x + p = 0 has real roots, its discriminant must be greater than or equal to zero. The discriminant is D = b^2 - 4ac = 8^2 - 4(5)(p) = 64 - 20p. Setting up the inequality 64 - 20p >= 0 and solving for p gives -20p >= -64. Dividing by -20 reverses the inequality, showing that p is less than or equal to 16/5.

Keep practicing — related questions