How to use the discriminant to find out what type of solutions the equation has for $x^{2} + 6 x + 5 = 0$ $x^2 + 6x + 5 = 0$ ?

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Answer 1 · 2015-05-18T21:42:32

$x^{2} + 6 x + 5$ $x^2+6x+5$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ with $a = 1$ $a=1$ , $b = 6$ $b=6$ and $c = 5$ $c=5$ .

The discriminant is given by the formula:

$Delta = b^2-4ac = 6^2-(4xx1xx5) = 36-20 = 16 = 4^2$

Since $Delta > 0$ , the quadratic equation $x^2+6x+5 = 0$ has two distinct real roots.

Further, since $Delta = 4^2$ is a perfect square the roots are rational.

(In fact the solutions are given by the formula:

$x = (-b+-sqrt(Delta))/(2a) = (-6+-4)/2 = -3+--2$

That is $x = -1$ or $x = -5$ .)

How to use the discriminant to find out what type of solutions the equation has for x^2 + 6x + 5 = 0x2+6x+5=0x^2 + 6x + 5 = 0?