# What are the integer solutions of the equation 4x^2-13x+3=0 ?

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4x^2 - 13x + 3

4x^2 - 12x - 1x + 3

group

(4x^2 - 12x) (- 1x + 3)

factor:

4x ( x - 3) -1(x - 3)

(4x - 1) (x-3)

set equal to 0

4x - 1 = 0

4x = 1

x = 1/4

x - 3 = 0

x = 3

You can do it as the answer above showed, or you can also solve by factoring:

4x^2 - 13x + 3

a b c

multiply a by c 4 x 3 =12

fiind factors of c that add up to b (-13) which would be -1 and -12

now plug those numbers in as b

4x^2 - 12x - 1x + 3

now group

(4x^2 - 12x) (- 1x + 3)

Now factor:

4x ( x - 3) -1(x - 3)

group the numbers outside of the parentheses:

(4x - 1) (x-3)

set them equal to 0 to find the solutions:

4x - 1 = 0

4x = 1

x = 1/4

x - 3 = 0

x = 3

To decide if the equation has an integer solution, we'll have to calculate them, first.

We'll apply quadratic formula to calculate the solutions of the equation:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x2 = [-b-sqrt(b^2 - 4ac)]/2a

a,b,c, are the coefficients of the quadratic

a = 4, b = -13 , c = 3

x1 = [13+sqrt(169 - 48)]/8

x1 = (13+11)/8

x1 = 24/8

x1 = 3

x2 = (13-11)/8

x2 = 2/8

x2 = 1/4

**Since x2 is not an integer number, we'll conclude that the only integer solution of the equation is x = 3.**