which equation has roots of 3 and 5

Answer 1

any equation that has these two factors set equal to zero, (x-3)(x-5) = 0

will have 3 and 5 as roots

x² -8x + 15 = 0 is the standard equation with those two roots or zeroes or solutions, that's what you get when you multiply (x-3) by (x-5). Just take the roots, change their sign and stick an x in front of them to get the factor.

but you could add in other factors, or just square the above and all would have at least two roots 3, and 5.

(x-100)(x-3)(x-5)=0 is a cubic equation with roots 100, 3 and 5. Add in other factors such

as (x-27)(x-100)(x-3)(x-5)=0 and you have a 4th degree equation with 4 roots including the 3 and 5.

You could also have an absolute value equation with those 2 roots

Take y=absolute value of x and shift it down and to the right, and the 2 perpendicular lines of the V shaped graph would intersect the x-axis at 3 and 5, as roots of the equation

Any equation that crossed the x-axis at 3 and 5 would have 3 and 5 as roots. An infinite number of equations could have those 2 roots.

Answer 2

Since the roots are x=3 and x=5, the factored form would be (x-3).(x-5) = 0

Multiplying the binomials yields x^2-8x+15 = 0