Question

For wt value of p , 2x2 + px + 8 =0 has real roots .

Answer 1

The quadratic formula says x = (-p +/- sqrt(p^2 - 4 * 2 * 8)) / (2*2). This yields nonreal solutions when p^2 - 4 * 2 * 8 < 0, real solutions otherwise. Thus, if the equation is to have real solutions, p^2 - 4 * 2 * 8 >= 0, or p^2 >= 64, or p >= 8.

Answer 2

Given,

• The quadratic equation $2x^2+px+8 = 0$ is given.
• The equation has real roots.

To find,

• The value of p

Solution,

For p = 8, $2x^2+px+8 = 0$ has real roots.

We can simply find the value of p if the equation has real roots by finding out the discriminant.

     D = b² - 4ac

In the quadratic equation,

     b = p

      a = 2

      c = 8

Substituting the values of a,b, and c in the formula of discriminant, we get

      D = (p)² - 4 (2) (8)

      D =   p² - 64

For the quadratic equation to have real roots, the value of the discriminant must be zero.

      D =0

   p² -64 = 0

         p² = 64

          p = 8

∴ The value of p = 8.

Hence, for p = 8, $2x^2+px+8 = 0$has real roots.