Question

From a set of first 10 natural numbers,three distinct prime numbers a,b,c are selected to form a quadratic equation ax^2+bx+c=0 having real roots.find the sum of the roots of all such possible quadratic equations that can be formed.

Answer 1

too tired

Answer 2

please try it..!if u can

Answer 3

is 0 included in "natural numbers"

Answer 4

possible prime numbers are 2, 3, 5, 7

Answer 5

yes

Answer 6

do you know how to find the roots? If so, use all possible combinations of 2,3,5,7 for a,b, and c , and find roots of all equations. add the roots of the equations with real roots together.

Answer 7

You can exclude some values of b 2 and 3 will never work out to real roots

Answer 8

Also for b = 5 and b = 7 only values or 2,3 for a and c will work. 4 quadratics

Answer 9

for roots to be real D >0.

Answer 10

that is b^2 - 4ac > 0

Answer 11

total 5 quadratic equations i think can be possible
where the sum of roots will be = -b/a

Answer 12

am I correct?

Answer 13

I only see 4 myself.

Answer 14

2x^2 + 5x + 3, 3x^2 + 5x + 2, 2x^2 + 7x + 3, 3x^2 + 7x + 2

Answer 15

Giving roots of -5/4+-1, -5/6+-1, -7/4+-5, -7/6+-5

Answer 16

when b=5 , a can take 2 or 3
when b=7 a can take 2,5,3

Answer 17

You're right, 2 and 5 work as a and c with 7. That gives two more equations.

Answer 18

-7/4+-3 and -7/6+-3

Answer 19

How come you arrive to a conclusion of -5/4+-1 as there roots
As I know sum of the roots will be -b/a
so it will be -5/2,-5/3,-7/2,-7/5,-7/3

Answer 20

That is my bad notation I mean -5/4(+-)1 as in -5/4 + 1 and -5/4 - 1

Answer 21

-5/4 + 1 and -5/4 - 1 ???? can you please explain?

Answer 22

That would come out -b/a for the sum of both.

Answer 23

ok.
Consider one of the quadratic equation being 2x^2+5x+3
The sum of roots in this case will be -5/2

Answer 24

correct?

Answer 25

Yes.

Answer 26

similarly you have 2x^2+5x+3,
3x^2+5x+2,
2x^2+7x+3,
2x^2+7x+5,
3x^2+7x+2,
3x^2+7x+5,
5x^2+7x+2,
5x^2+7x+3,

Answer 27

So you'll have what you wrote above and -7/2 again since 2x^2 + 7x + 3 and 2x^2 + 7x + 5 both give -7/2

Answer 28

49 < 4*5*3

Answer 29

So not 5x^2 + 7x + 3 or 3x^2 + 7x + 5

Answer 30

Otherwise you have them all.

Answer 31

yes in that case D< 0 and roots will be imaginary

Answer 32

So finally we have
2x^2+5x+3, 3x^2+5x+2, 2x^2+7x+3, 2x^2+7x+5, 3x^2+7x+2,5x^2+7x+2

Answer 33

That is it.

Answer 34

I get 149/10 out of mathematica as the final sum

Answer 35

so it will be
-5/2-5/3-7/2-7/2-7/3-7/5 = -149/10

Answer 36

Yes.

Answer 37

Correct...Thanks!!!!I appreciate it.

Answer 38

No problem