Question

Find m such that mx^2/2 + (m-3)x + 2 = 0 has equal roots. Show working out thanks.

Answer 1

"has equal roots"
is perhaps supposed to read
"has two equal roots"
?

Answer 2

nope, on my test is says has equal roots but it means the same thing.

Answer 3

it*

Answer 4

well in any event what part if a quadratic determines what kind of roots it has?

Answer 5

it's discriminant value? like if it's >0 then it has two real solutions, =0 it has one solution and <0 it has no real solutions.

Answer 6

right, and what is the discriminant here?

Answer 7

(m-3)^2 - 4(1/2m)(2)

Answer 8

yes, and what do we set it equal to

Answer 9

?

Answer 10

0?

Answer 11

good, now solve for m

Answer 12

I actually already knew and solved it already, but I just wanted to understand it because I am still a bit confused as why do you use the discrimant to find the roots? and why is there the additional m infront of the x^2 and what does it represent?

Answer 13

the m is number that determines the coefficients in the expression
we want to find m such that we get a double root from the quadratic
the reason the discriminant is what tells us about out roots id because of how it appears int eh quadratic formula\[x={-b\pm\sqrt{b^2-4ac}\over2a}\]if \(b^2-4ac>0\) then the number under the radical is real, and so we get two different answers for x
(it's two answers because of the \(\pm\) part)

Answer 14

if \(b^2-4ac<0\) then we wind up taking the square root of a negative number in the quadratic formula, which gives an imaginary number, and so we get two complex answers.
(two answer, again, due to the \(\pm\) part)

Answer 15

ok i think i get it now, thanks!

Answer 16

if \(b^2-4ac=0\) then the square root of that is also zero, and we are left with\[x=\frac{-b}{2a}\]which we imagine occurs twice to make up for the fact that we have an invisible\(\pm0\) next to the \(-b\)

Answer 17

you're welcome :)