Question

solve for x. 4x^2=-13x-3

Answer 1

rewrite the problem as

\[4x^2 + 13x + 3 = 0\]
'
so to solve this I multiply 4 and 3 then find the factors that add to 13

any ideas..?

Answer 2

12 and 1

Answer 3

great so next write the factorisation as

\[\frac{(4x + 12)(4x + 1)}{4}\]

then remove the common factors from the numerators \[\frac{4(x + 3)(4x + 1)}{4} = 0\]

cancel out the common factor in the numerator ad denominator

then you can find the solutions

Answer 4

so (x+3)(4x+1)

Answer 5

that's the factored for so you have

x +3 = 0 or 4x + 1 = 0

so find the 2 values of x that make the equations true

Answer 6

-3 and -1/4

Answer 7

that's correct