2 answers

How do you solve #3x^2+13x=-4#?

Question:

How do you solve #3x^2+13x=-4#?

Answers

See a solution process below:

Explanation:

First, add #color(red)(4)# to each side of the equation to put the equation in standard form:

#3x^2 + 13x + color(red)(4) = -4 + color(red)(4)#

#3x^2 + 13x + 4 = 0#

Next, we can factor the left side of the equation as:

#(3x + 1)(x + 4) = 0#

Now, solve each term on the left for #0# to find the solutions for #x#:

Solution 1:

#3x + 1 = 0#

#3x + 1 - color(red)(1) = 0 - color(red)(1)#

#3x + 0 = -1#

#3x = -1#

#(3x)/color(red)(3) = -1/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -1/3#

#x = -1/3#

Solution 2:

#x + 4 = 0#

#x + 4 - color(red)(4) = 0 - color(red)(4)#

#x + 0 = -4#

#x = -4#

The Solutions Are: #x = -1/3# and #x = -4#

.

#x=-4" or "x=-1/3#

Explanation:

#"rearrange and equate to zero"#

#rArr3x^2+13x+4=0#

#"the factors of the product of 12 which sum to 13 are"#
#"12 and 1"#

#"split the x-term and factor by grouping"#

#3x^2+12x+x+4=0#

#color(red)(3x)(x+4)color(red)(+1)(x+4)=0#

#"factor out "(x+4)#

#rArr(x+4)(color(red)(3x+1))=0#

#"equate each factor to zero and solve for x"#

#x+4=0rArrx=-4#

#3x+1=0rArrx=-1/3#

.

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