Write the quadratic equation whose roots are
−4 and 6, and whose leading coefficient is 5.
y=5(x+4)(x-6)
is y=5(x+4)(x-6) the final answer or do i have to do anything else to that
Yes, I get the same answer! You were not asked to solve for x. f(x) = 5(x-6)(x+4)
To write the quadratic equation with roots −4 and 6, and a leading coefficient of 5, we can use the fact that the roots of a quadratic equation in the form of ax^2 + bx + c = 0 are given by the formula:
x = (-b ± sqrt(b^2 - 4ac))/(2a)
Given that the roots of the quadratic equation are −4 and 6, let's substitute them into the formula:
For the root x = -4:
-4 = (-b ± sqrt(b^2 - 4ac))/(2a)
For the root x = 6:
6 = (-b ± sqrt(b^2 - 4ac))/(2a)
Since the leading coefficient is 5, we can substitute a = 5:
For the root x = -4:
-4 = (-b ± sqrt(b^2 - 20c))/10
For the root x = 6:
6 = (-b ± sqrt(b^2 - 20c))/10
Now, we can solve these two equations simultaneously to find the values of b and c.
From the equation -4 = (-b ± sqrt(b^2 - 20c))/10, we can simplify it to -40 = -b ± sqrt(b^2 - 20c).
Similarly, we can simplify the equation 6 = (-b ± sqrt(b^2 - 20c))/10 to 60 = -b ± sqrt(b^2 - 20c).
Now, we have a system of equations:
-40 = -b ± sqrt(b^2 - 20c)
60 = -b ± sqrt(b^2 - 20c)
To find the values of b and c, we can eliminate the square roots by squaring both sides of the equations:
(-40)^2 = (-b ± sqrt(b^2 - 20c))^2
(60)^2 = (-b ± sqrt(b^2 - 20c))^2
Simplifying further:
1600 = b^2 - 20c (1)
3600 = b^2 - 20c (2)
From equations (1) and (2), we can see that they are both the same. This is because squaring a positive number and squaring a negative number will give the same result.
Therefore, we can conclude that 1600 = 3600.
Since this equation cannot be true, it means that there is no quadratic equation with the given roots and leading coefficient.