What is the discriminant of $d^2− 7d + 8 = 0$ and what does that mean?

Question

2696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-28T16:22:20

For this quadratic, $Delta = 17$ , which means that the equation has two distinct real roots.

For a quadratic equation written in the general form

$ax^2 + bx + c = 0$

the determinant is equal to

$Delta = b^2 - 4 * a * c$

Your quadratic looks like this

$d^2 - 7d + 8 = 0$ ,

which means that, in your case,

${(a=1), (b = -7), (c = 8) :}$

The determinant for your equation will thus be equal to

$Delta = (-7)^2 - 4 * (1) * (8)$

$Delta = 49 - 32 = color(green)(17)$

When $Delta>0$ , the quadratic will have two distinct real roots of the general form

$x_(1,2) = (-b +- sqrt(Delta))/(2a)$

Because the discriminant is not a perfect square, the two roots will be irrational numbers.

In your case, these two roots will be

$d_(1,2) = (-(-7) +- sqrt(17))/(2 * 1) = {(d_1 = 7/2 + sqrt(17)/2), (d_2 = 7/2 - sqrt(17)/2) :}$

What is the discriminant of d^2− 7d + 8 = 0d^2− 7d + 8 = 0 and what does that mean?