By what amount does the sum of the roots exceed the product of the roots of the equation (x-7)(x+3)=0?

I have no idea what to do!

first get the roots,,

(x-7)(x+3)=0
thus equating each term inside the parenthesis to zero,
x-7=0 and x+3=0
therefore, x=7 and x=-3
the roots are 7 and -3
sum: 7 + (-3) = 4
product: 7(-3) = -21
the questions asks for the difference between the sum and product, therefore:
4 - (-21) = 25

hope this helps~ :)

To find the sum and product of the roots of the given equation (x-7)(x+3)=0, we first need to find the roots.

Setting each factor equal to zero and solving for x, we get:

x - 7 = 0 --> x = 7
x + 3 = 0 --> x = -3

So the roots of the equation are x = 7 and x = -3.

To find the sum of the roots, we add them together:

Sum = 7 + (-3) = 4

To find the product of the roots, we multiply them together:

Product = 7 * (-3) = -21

Now, we can calculate by how much the sum of the roots exceeds the product of the roots:

4 - (-21) = 4 + 21 = 25

Therefore, the sum of the roots exceeds the product of the roots by 25.

Hope this helps! Let me know if you have any further questions.