Discriminant ProblemsWe all know that the discriminant is the part b2−4ac of the equationx=−b±b2−4ac2athat we use to find the roots of a quadratic equation eg:ax2+bx+c=0 or the part in a trinomial expression like ax2+bx+c.What I want to know is what kind of ideas we can conclude for a given inequality of a discriminant. We know that in quadratic formulas the number of real roots depends on the case whether the discriminant is &lt;0, 0, or &gt;0. But if given only the discriminant and the inequality of it, for examples like this:b2−4ac=0b2−4ac&gt;0b2−4ac&lt;0What can we conclude by the given information for each case? How can it help to reveal certain problems involving trinomial expressions?

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Discriminant ProblemsWe all know that the discriminant is the part b2−4ac of the equationx=−b±b2−4ac2athat we use to find the roots of a quadratic equation eg:ax2+bx+c=0 or the part in a trinomial expression like ax2+bx+c.What I want to know is what kind of ideas we can conclude for a given inequality of a discriminant. We know that in quadratic formulas the number of real roots depends on the case whether the discriminant is &amp;lt;0, 0, or &amp;gt;0. But if given only the discriminant and the inequality of it, for examples like this:b2−4ac=0b2−4ac&amp;gt;0b2−4ac&amp;lt;0What can we conclude by the given information for each case? How can it help to reveal certain problems involving trinomial expressions?

srasloavfv · Accepted Answer

Let me rewrite the quadratic equation asx=−b2a±−1a(4ac−b24a).Notice that I wrote it in terms of the coordinates of the vertex. Now, let us consider some cases about the vertex&amp;nbsp;(h,k), assuming that the quadratic opens upward.1)&amp;nbsp;k&amp;lt;0In this case, the vertex is below the&amp;nbsp;x-axis and the quadratic must have two real roots. Now,4ac−b24a&amp;lt;04ac−b2&amp;lt;0−(4ac−b2)&amp;gt;0b2−4ac&amp;gt;02)&amp;nbsp;k=0This means that the vertex is on the&amp;nbsp;x-axis and the quadratic has a double root. Now, making a similar solution from the first case,4ac−b24a2=04ac−b2=0−4ac+b2=0b2−4ac=0This matches our criteria that a quadratic has a double root if and only if&amp;nbsp;b2−4ac=0.3)&amp;nbsp;k&amp;gt;0.In this case, the vertex is above the&amp;nbsp;x-axis and since it is the minimum of the function, the quadratic does not have any real root. Now,4ac−b24a&amp;gt;04ac−b2&amp;gt;0−4ac+b2&amp;lt;0b2−4ac&amp;lt;0and this matches our criteria that a quadratic does not have any real root if and only if&amp;nbsp;b2−4ac&amp;lt;0.

Discriminant Problems We all know that the discriminant is

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