Question

Another Question based on Nature of roots

Answer 1

If the roots of the equation x^2 -2cx +ab =0 are real and unequal, then prove that the roots of x^2 -2(a+b)x +a^2 +b^2 +2c^2 =0 will be imaginary

Answer 2

there is an easy way and an hard way to do this

Answer 3

I got c^2 -ab >0

Answer 4

I'll give u the easy way first may be :)

Answer 5

yes

Answer 6

Easy way :

\(c^2 -ab >0 \implies c^2 \gt ab\)

\[ x^2 -2(a+b)x +a^2 +b^2 +2c^2 \gt x^2 -2(a+b)x +a^2 +b^2 +2ab \]
\[ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \gt x^2 -2(a+b)x +(a+b)^2 \]
\[ ~~~~~~~~~~~~~~~~~\gt (x-(a+b))^2 \]
\[ ~ \gt 0 \]


QED

Answer 7

don't give the hard way then
Maybe i will figure it out later

Answer 8

Okay, the hard way is straight forward but will be painful and you wont learn anything from that method as all u do is just arithmetic and algebra manipulaiton

Answer 9

thank u !

Answer 10

yw

Answer 11

x2−2(a+b)x+a2+b2+2c2>x2−2(a+b)x+a2+b2+2ab
Can u please explain the second step , I mean what u did here

Answer 12

I have just replaced c^2 wid ab

Answer 13

We have : \(\large c^2 \gt ab\)

So, \(\large 2c^2 \gt 2ab\)

Answer 14

And the
(x−(a+b))2 >0

Answer 15

used the identity : \(\large u^2 -2uv + v^2 = (u-v)^2\)
\(u = x, \\ v = a+b\)

Answer 16

Not about that
If it > 0 how it is proved

Answer 17

thats a very good question !

Answer 18

let me ask u a question first :
How does the graph of a quadratic look when the roots are imaginary ?