Question

what is the nature of the roots of the equation 5n^2=4n+6?

Answer 1

first write it as 5n^2 - 4n - 6 = 0

thats in standard form

ax^2 + bx + c = 0

the nature of its roots is found by working out the value of the discriminant
which is b^2 - 4ac where a,b and c are the values in the standard form

Answer 2

so if you compare 5n^2 - 4n - 6 = 0 with the standard form
a = 5 , b = -4 and c = -6

right?

do you follow that? ( ignore difference in the variables x and n )

Answer 3

yes i follow :)

Answer 4

so you have b^2 - 4ac = (-4)^2 - 4*5*-6

if this is > 0 there are 2different real roots
i this = 0 there are 2 duplicate real roots
if < 0 there are no real roots

Answer 5

yes I have that. should my answer be 136?

Answer 6

yep - so how many roots are there and are they real?

Answer 7

i'm not sure about roots -_-

Answer 8

the reason we use the discriminant is because its part of the formula for roots of a quadratic equation and its square roots is taken


x = [-b +/- sqrt(b^2-4ac)] / 2a

Answer 9

the roots of an equation are also called the zeros

Answer 10

the answer here is 2 real zeros because the discriminant is = 136 which is > 0

Answer 11

Ohh okay! That's what I wasn't understanding, sorry. I'm trying to study for an exam that's tomorrow