Can someone solve this....?

Question

Using quadratic equation or whatsoever:
Without solving, find the sum and product of the solutions of 6y^2-5x+3=0......y^2 means y squared

Another question:
1.) Find a quadratic equation for which the sum of the solutions is 2/3 and the product of the solution is 5/6

2.) Approximate solutions of 2x^2-5x+3=0.....x^2 mean x squared

uhhh..... for no.2 I mean approximate solutions to the nearest tenth

Answer 1

Well, it says "without solving", so that means no quadratic equation for you.

6y^2 - 5y + 3 = 0
<=> y^2 - (5/6) y + (1/2) = 0

Say solutions are a and b, then this is the same as
(y-a)(y-b) = 0
<=> y^2 - (a+b) y + ab = 0

Comparing these we have
a+b = 5/6
ab = 1/2
So the sum of the solutions is 5/6 and their product is 1/2.

Edit: Just noticed your new questions.
1.) Find a quadratic equation for which the sum of the solutions is 2/3 and the product of the solution is 5/6

As above, we can write this down directly as
x^2 - 2/3 x + 5/6 = 0
and we can make it a little nicer by multiplying by 6:
6x^2 - 4x + 5 = 0.

2.) Approximate solutions of 2x^2-5x+3=0
Not sure why you want approximate solutions, since it's easy to factorise:
2x^2-5x+3=0
<=> (2x - 3) (x - 1) = 0
<=> x = 1 or 3/2.

Answer 2

assuming that x and y are orthogonal then the solutions of 6y^2-5x+3=0 lie on two infinitely long parallel lines if you can find a way to take the sum or product of these lines then you are ein chenius!

Answer 3

original eqn: 6y2 -5x +3=0
=> x=(6y2)/5 + 3/5

now its a simple quadratic eqn of form
x=ay2 + by + c
whose sum of roots is -b/a
product is c/a

hence the sum is 0 as no b here
product = 1/2