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The solutions of the quadratic equation Ax^2+Bx+C=0 are called roots and are given by a famous equation called the Quadratic Formula . In the quadratic formula, the terms B^2-4AC underneath the radical have a special name: the Discriminant .

Discuss the relationship between the discriminant of a quadratic polynomial and the quantity of real roots it possesses. Explain the positioning of the roots of the polynomial on its graph with respect to the discriminant and the sign of the discriminant.


Sagot :

The roots of a quadratic equation depends on the discriminant [tex]\Delta[/tex].

  • If [tex]\Delta > 0[/tex], the quadratic equation has two real distinct roots, and it crosses the x-axis twice.
  • If [tex]\Delta = 0[/tex], the quadratic equation has one real root, and it touches the x-axis.
  • If [tex]\Delta < 0[/tex], the quadratic equation has two complex roots, and it neither crosses nor touches the x-axis.

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A quadratic equation has the following format:

[tex]y = ax^2 + bx + c[/tex]

It's roots are:

[tex]y = 0[/tex]

Thus

[tex]ax^2 + bx + c = 0[/tex]

They are given by:

[tex]\Delta = b^{2} - 4ac[/tex]

[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]

The discriminant is [tex]\Delta[/tex].

  • If it is positive, [tex]-b + \sqrt{\Delta} \neq -b - \sqrt{\Delta}[/tex], and thus, the quadratic equation has two real distinct roots, and it crosses the x-axis twice.
  • If it is zero, [tex]-b + \sqrt{\Delta} = -b - \sqrt{\Delta}[/tex], and thus, it has one real root, and touching the x-axis.
  • If it is negative, [tex]\sqrt{\Delta}[/tex] is a complex number, and thus, the roots will be complex and will not touch the x-axis.

A similar problem is given at https://brainly.com/question/19776811