finding the zeros in a function: h(x) = 6x^2 - 7x + 2

[jferrel8]

h(x)= 6x[2]-7x+2
(^ that is six "X" squared)

My work (can you check this to see if i am correct...thanks!):

A= 6
B= -7
C= 2 (using the quadratic formula i got) = 7 +-/(-7)[2]-4(6)(2) all over 12... therfore 7+- /1 all over 12
(^ all within a square root) (^ square root)

then i got 7+1/12= 2/3 and 7-1/12=2

 

[Deleted member 4993]

Re: finding the zeros in a function

h(x)= 6x[2]-7x+2
(^ that is six "X" squared)

My work (can you check this to see if i am correct...thanks!):

A= 6
B= -7
C= 2 (using the quadratic formula i got) = 7 +-/(-7)[2]-4(6)(2) all over 12... therfore 7+- /1 all over 12
(^ all within a square root) (^ square root)

then i got 7+1/12= 2/3 and 7-1/12=2

To check your work see if h(2/3) = 0 and h(2) = 0

 

[stapel]

h(x)= 6x[2]-7x+2
(^ that is six "X" squared)...

...(using the quadratic formula i got) = 7 +-/(-7)[2]-4(6)(2) all over 12... therfore 7+- /1 all over 12
(^ all within a square root) (^ square root)

As you noticed when you reviewed what you'd posted, the forum script does not "respect" blank spaces. To format your work clearly, kindly use the formatting explained in the articles at the link you saw in the "Read Before Posting" thread. Thus, in typed characters, as:

. . . . .h(x) = 6x^2 - 7x + 2

. . . . .x = (7 +/- sqrt[(-7)^2 - 4(6)(2)]) / ((2)(6))

...or, in LaTeX, as:

. . . . .\(\displaystyle h(x)\, =\, 6x^2\, -\, 7x\, +\, 2\)

. . . . .\(\displaystyle x\, =\, \frac{7\, \pm\, \sqrt{(-7)^2\, -\, 4(6)(2)}}{(2)(6)}\)

...and so forth.

Thank you!

Eliz.

 

[jferrel8]

thank you!