Choose the quadratic equation that has a leading coefficient of 1 and solutions 3 and -2.
options:
a.) x^2+x+6=0
b.) x^2-x-5=0
c.) x^2+x+5=0
d.) x^2-x-6=0

Option D

Step-by-step explanation:

Verify each quadratic equation

case A) we have

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

----> is not true

therefore

x=3 is not a solution of the quadratic equation

case B) we have

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

----> is not true

therefore

x=3 is not a solution of the quadratic equation

case C) we have

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

----> is not true

therefore

x=3 is not a solution of the quadratic equation

case D) we have

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

----> is true

therefore

x=3 is a solution of the quadratic equation

For x=-2

----> is true

therefore

x=-2 is a solution of the quadratic equation

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

the answer is d. (1.2, -4.7) on edgn

hope this !