Quadratic equation whose roots are 5 and -5 with coefficient 3

using the letter x to represent the variable

please show steps

What did you get ?

3(x-5) (x+5)= 3x^2-28x + 75

Your expansion is incorrect

3(x-5)(x+5)
= 3(x^2 - 25)
= 3x^2 - 75

Since you wanted a quadratic equation, it should be

3x^2 - 75 = 0

To find the quadratic equation with roots 5 and -5, we can use the fact that the roots of a quadratic equation with the form ax^2 + bx + c = 0 are given by the equations:

x = (-b ± √(b^2 - 4ac))/(2a)

Since the roots are 5 and -5, we can substitute these values into the equation:

5 = (-b ± √(b^2 - 4ac))/(2a)
-5 = (-b ± √(b^2 - 4ac))/(2a)

Now, let's solve for b, the coefficient of the linear term, a, the coefficient of the quadratic term, and c, the constant term.

Using the first equation, where x = 5:

5 = (-b ± √(b^2 - 4ac))/(2a)

Simplifying, we have:

5 = (-b ± √(b^2 - 4ac))/(2a)
10a = -2b ± √(b^2 - 4ac)
10a + 2b = ± √(b^2 - 4ac)
(10a + 2b)^2 = b^2 - 4ac

Expanding and simplifying:

100a^2 + 40ab + 4b^2 = b^2 - 4ac
100a^2 + 40ab = -4ac

Dividing both sides by 4 gives us:

25a^2 + 10ab = -ac

Dividing through by a and simplifying:

25a + 10b = -c

Using the second equation, where x = -5, we repeat the same steps:

-5 = (-b ± √(b^2 - 4ac))/(2a)

Simplifying:

-5 = (-b ± √(b^2 - 4ac))/(2a)
-10a = -2b ± √(b^2 - 4ac)
-10a + 2b = ± √(b^2 - 4ac)
(-10a + 2b)^2 = b^2 - 4ac

Expanding and simplifying:

100a^2 - 40ab + 4b^2 = b^2 - 4ac
100a^2 - 40ab = -4ac

Dividing both sides by 4:

25a^2 - 10ab = -ac

Dividing through by a:

25a - 10b = -c

Now, we have two equations:

25a + 10b = -c
25a - 10b = -c

We can add these equations together to eliminate the c terms:

(25a + 10b) + (25a - 10b) = -c + (-c)
50a = -2c

Dividing through by -2 gives us:

-25a = c

So, we have found that the constant term, c, is equal to -25a.

Now, let's substitute this back into one of the original equations, for example, 25a + 10b = -c:

25a + 10b = -(-25a)
25a + 10b = 25a
10b = 0

This means that b must be equal to 0.

Finally, we substitute a = 3 (as given in the question) into the equation -25a = c:

-25(3) = c
-75 = c

Therefore, the quadratic equation with roots 5 and -5 and coefficient 3 is:

3x^2 + 0x - 75 = 0