Question

solve what are the solutions and find the x-intercepts : 7x+x(x-4)=0
solve what are the solutions and find the x-intercepts : 7x+x(x-4)=0
@Mathematics

Answer 1

\[x^2-4x+7x=0\]\[x^2+3x=0\]\[x^2+3x+\frac{9}{4}=\frac{9}{4}\]\[\left(x+\frac{3}{2}\right)^2=\frac{9}{4}\]\[x+\frac{3}{2}=\pm\frac{3}{2}\]\[x=0,3\]

Answer 2

x = 0 and x = 3

Answer 3

Ah across gotcha.

Answer 4

The solution set is {-3, 0} (x=3 is not a solution.)

Answer 5

Mandolino is right. Good observation.

Answer 6

@across, when you got to:\[x^2+3x=0\]this could have been simplified to:\[x(x+3)=0\]so x=0 or -3

Answer 7

Another method: you could factor out an x
7x+x(x-4)=0
x(7+x-4)=0
x(x+3)=0
x=0 or x+3=0
x=-3
Just for the sake of interest :)

Answer 8

yeah I mess that up, x does not equal 3.

Answer 9

so the solutions are 0 and 3 and the x intercepts would be 3,0 and 0,0???

Answer 10

No, the solutions are 0 and -3

Answer 11

o thanks i see it now

Answer 12

The solutions to the equation set equal to 0 are also the x-intercepts if we think of it as a function.

Answer 13

does anyone know how to find and label the vertex and line of symmetry?