Question

Which of the following are solutions to the equation below?

Check all that apply.

3x^2 + 27x + 60 = 0

A. 4
B. –4
C. –5
D. 5
E. –27

Answer 1

B. -4 and C.-5
3x^2+27x+60=0 divide by 3
3(x^2+9x+20)=0 simplify
3(x+4)(x+5)=0
Use the 0 property and check your work.

Answer 2

The solutions are B. -4 and C. -5

Explanation:

For a quadratic equation of the form
`ax^2+bx+c=0` the solutions are

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

For
`\mathrm{}\quad a=3,\:b=27,\:c=60:\quad x_(1,\:2)=(-27\pm √(27^2-4\cdot \:3\cdot \:60))/(2\cdot \:3)`

`x_1=(-27+√(27^2-4\cdot \:3\cdot \:60))/(2\cdot \:3)\\\\x_1=(-27+√(9))/(2\cdot \:3)\\\\x_1=(-27+3)/(2\cdot \:3)\\\\x_1=(-24)/(6) = -4`

`x_2=(-27-√(27^2-4\cdot \:3\cdot \:60))/(2\cdot \:3)\\\\x_2=(-27-√(9))/(2\cdot \:3)\\\\x_2=(-27-3)/(2\cdot \:3)\\\\x_2=-(30)/(6) = -5`