How do you solve #3x^2+5x+2=0 #?

Question

Savannah Anderson · Accepted Answer

#(3x+2)(x+1)#

Since both 3 and 2 are prime, their only factors are 1 and 3 in the case of 3 and 1 and 2 in the case of 2.

Assuming there are integer solutions for #x# then it has to be true that one of the following is true. If not, then it is down to the formula.

#(3x+-2)(x+-1) # as the first option #(3x+-1)(x+-2)# as the second option

In the given question: The constant 2 is positive so the signs in the brackets are the same. The #5x# is positive so both signs are +

#color(blue)("Test 1")#

#(3x+2)(x+1)#

# =" " 3x^2+3x+2x+2#

#color(blue)( =" " 3x^2+5x+2)#

Jace Anderson · Answer

undefined To solve the quadratic equation 3x^2 + 5x + 2 = 0, you can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 3x^2 + 5x + 2 = 0, where a = 3, b = 5, and c = 2, plug these values into the quadratic formula:

x = (-5 ± √(5^2 - 4 * 3 * 2)) / (2 * 3)

Calculate the discriminant (b^2 - 4ac):

b^2 - 4ac = 5^2 - 4 * 3 * 2 = 25 - 24 = 1

Now substitute the discriminant into the formula:

x = (-5 ± √1) / 6

Since the square root of 1 is 1:

x = (-5 ± 1) / 6

This yields two solutions:

x₁ = (-5 + 1) / 6 = -4/6 = -2/3
x₂ = (-5 - 1) / 6 = -6/6 = -1

So, the solutions to the equation 3x^2 + 5x + 2 = 0 are x = -2/3 and x = -1.