Write the quadratic equation whose roots are 2 and -5 and whose leading confessions is 3

Answer 1

If roots are 2 and -5 that means that we have the factors too!

x=2 x=-5

x-2=0 x+5=0

Go back into zero-factor property

(x-2) (x+5) = 0

Now add the leading coefficient to it.

3(x-2) (x+5) = 0

distribute the 3 into the factor (x-2)

(3x-6) (x+5)

Multiply the factors

3x²+15x-6x-30

Combine like terms and then you will get the quadratic equation

3x²+9x-30

Answer 2

Jose, you probably noticed the two answers of Andrea and Martin are different. They treated the initial coefficient of 3 differently. So they aren’t both right.

When the tutors have different answers, it’s wise to check their work. Since we were given roots of 2 and -5, you can test by substituting 2 for x in each formula, and do the same for -5.

Answer 3

The roots of a quadratic equation are the x values where the curve crosses the x axis. Therefore, at those points y is equal to zero, So using the roots, 2 and -5, create two binomials such that subtracting the roots will have a result of zero. For these roots, your binomials will be (x - 2) and ((x + 5). Another way to think of it is subtract the roots from x to get the binomials.

Now multiply thos binomials together:

(x - 2) ((x + 5) = x² + 5x - 2x -10, combine terms:

x² +3x -10,

Now put in the lead coefficient. By definition, the lead coefficient multiplies the variable raised to the highest power, in this case, x². So the final equation becomes:

y = 3x² + 3x -10

Hope this helps