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

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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) :}#