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

givingiswinning | Student, Grade 10 | (Level 1) Valedictorian

Posted on

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

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.