Question

Show that the roots of the equation x^2+2(3a+5)x+2(9a^2+25)=0 are complex unless a=5/3.

Answer 1

first step: Evaluate Discriminant

Answer 2

and of course we know when discriminant of a quadratic is negative then roots are complex

Answer 3

solving discriminant fully i get 12a(-3a+10)

Answer 4

@Hero help

Answer 5

check ur work again

Answer 6

in the first step i get :
(2(3a+5))^2-4(2(9a^2+5))

Answer 7

should I multiply completely after this ?

Answer 8

What @mukushla is telling you is accurate.

Answer 9

well how's that?

Answer 10

how can we say that..? and how can we say that it is complex unless only a=5/3

Answer 11

yeah i know when discriminant is negative it is complex but how to show that? and that exception of a=5/3

Answer 12

tnx hero

just simplfy that expression completely and see what happens

Answer 13

ok. lemme solve.

Answer 14

-36a^2-100+120a

Answer 15

oh sorry its ok

Answer 16

-36a^2-100+120a

Answer 17

yeah that's it what i get.

Answer 18

\[-36a^2+120a-100=-4(9a^2-30a+25)\]

Answer 19

yup!

Answer 20

but it shouldn't mean that −4(9a2−30a+25) is negative.?!

Answer 21

because this term " (9a2−30a+25)" may also be negative

Answer 22

9a^2=(3a)^2
25=5^2

Answer 23

I'm pretty sure that whole thing is supposed to be set equal to zero.

Answer 24

how this? 9a^2=(3a)^2
25=5^2 ??

Answer 25

@Hero if we set that equal to zero it would mean that roots are real and equal.

Answer 26

because the discriminant would be zero.

Answer 27

see coolshubhs\[-4(9a^2−30a+25)=-4(3a-5)^2\]this is negative unless\[-4(3a-5)^2=0\]or\[a=\frac{5}{3}\]

Answer 28

yes .. you got this but i dont understand why you set this expression equal to zero?

Answer 29

well thanks dude :)