<p>If you take the discriminant of a quadratic equation, and you find that it is a perfect square. Are the roots, two real rational equal roots?</p>
<p>Also if the discriminant is equal to zero, is it true that there is only one root? </p>
<p>Can anyone elaborate on both of these examples?</p>
<p>not equal roots, but they’ll be rational. The only square root in the quadratic equations is in the square root of the discriminant…so if you square root a perfect square, you take out the only way to get an irrational number.</p>
<p>If the discriminant is zero in the quadratic equation, then you dont need even need to consider it. Then, the root is -b/2a, instead of (-b±sqrt(b^2-4ac))/2a.</p>
<p>so if i understand correctly, if the discriminant is greater than 0, the roots are unequal, real, irrational roots. </p>
<p>If the D is equal to 0, then the roots are equal, rational, and real</p>
<p>if the D is less than 0, then the roots are irrational. </p>
<p>Is all of this true?</p>
<p>kind of…the square root of a non-perfect square or a negative number(ie, the discriminant) is the only thing that can give you an irrational root. If the discriminant is greater than 0 and a non-perfect square, you’ll have unequal, irrational roots, but if it’s a perfect square, you’ll have rational roots(though not necessarily equal). You’ve got the rest right.</p>
<p>Not quite. If the discriminant is negative, then you don’t have any real roots, rational or irrational. This is because the roots contain the square root of the discriminant, which is imaginary if the discriminant is negative.</p>
<p>Wikipedia sums it up nicely:</p>
<p>For instance, the quadratic polynomial P(x) = ax^2 + bx + c has discriminant D = b^2 − 4ac, which is the quantity under the square root sign in the quadratic formula. For real numbers a, b, c, one has: </p>
<p>When D > 0 , P(x) has two distinct real roots x = (−b ± √D) / 2a , and its graph crosses the x-axis twice.
When D = 0, P(x) has exactly one real root x = −b / 2a, and its graph is tangent to the x-axis.
When D < 0 , P(x) has no real roots, and its graph lies strictly above or below the x-axis. </p>
<hr>
<p>Also, the statement that “if it’s a perfect square, you’ll have rational roots” is true only if b and a are rational. If, for example, a=pi, b=8 and c=1/pi, then D=4, but the roots aren’t rational.</p>
<p>I have one more question on this topic. If the discriminant is zero, are the roots rational or are they irrational?</p>
<p>Blah. I counted roots with i as still irrational…forgot to talk about the part about them not being real ;p</p>
<p>zero = roots are rational and the same.</p>
<p>thanks everyone</p>
<p><<zero ==“” roots=“” are=“” rational=“” and=“” the=“” same.=“”>> </zero></p>
<p>This is true only if:</p>
<ol>
<li><p>a and b are both rational; or</p></li>
<li><p>a and b are both irrational and one is a rational multiple of the other.</p></li>
</ol>
<p>Also, roots with i are not necessarily irrational. 2i is not irrational, and is the (double) root of x^2+2=0.</p>