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

Respuesta :

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:

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

Step-by-step explanation:

For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are

[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

For [tex]\mathrm{}\quad a=3,\:b=27,\:c=60:\quad x_{1,\:2}=\frac{-27\pm \sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}[/tex]

[tex]x_1=\frac{-27+\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_1=\frac{-27+\sqrt{9}}{2\cdot \:3}\\\\x_1=\frac{-27+3}{2\cdot \:3}\\\\x_1=\frac{-24}{6} = -4[/tex]

[tex]x_2=\frac{-27-\sqrt{27^2-4\cdot \:3\cdot \:60}}{2\cdot \:3}\\\\x_2=\frac{-27-\sqrt{9}}{2\cdot \:3}\\\\x_2=\frac{-27-3}{2\cdot \:3}\\\\x_2=-\frac{30}{6} = -5[/tex]