solve the equation y^2-4y+3=0?

Answer 1

(y-3)(y-1)=0
y=3 y=1

Answer 2

y^2-4y+3=0?

This is a good one to start with because it’s really rather trivial.
y² - 4y + 3 = 0

You can almost be certain that the factors will look like
(y + a)(y + b)
That’s because (y)(y)=y²
Multiplying that out, you get
y² + (a+b)y + ab

Let’s look at a and b…
ab=3 That means that they are 1 and 3…. Maybe

But you also know that a+b= -4. But 1+3=4, not -4. However, not all is lost. (-1)(-3)=3 also. And -1 + -3 = -4

So the factors are
(y + -1)(y + -3), or, if you prefer…
(y – 1)(y – 3)

Answer 3

[y-1][y-3]=0,so,

y-1=0, OR y-3=0,so,

y=1, OR y=3.

Answer 4

y^2 - 4 y + 3 = 0.

Always try to FACTOR first. The factors of 3 are 1 and 3, same sign (since it's +3), and notice the middle term is NEGATIVE. So we try:

(y-3)(y-1) = 0, which, you can check, works.

This means

y-3 = 0 | y-1 =0 (bar means OR)
y=3 | y=1

Done.

Answer 5

factor out y^2 -4y=3=0
(y-3)(y-1) open and close parenthesis means that you have to multiply using the FOIL method for checking it to the org. eq.
y-3=0 transpose 3 to the right side = +3;so y=3
y-1=0 " " " " " side=Y =1
DONE....

Answer 6

Y^2 - 4Y + 3 = 0
(Y - 3) (Y - 1) = 0
Y = 3
Y = 1