Question

Find the x-intercepts (if any) for the graph of the quadratic function.? `6x^2 +12x+5=0`

Give your answers in exact form. Show your work

Answer 1

Just apply the formula `x=(-b(+)or(-)(b^2-4*a*c)^(1/2))/(2*a)`
where the quadratic function is `a*x^2+b*x+c=0`

Explanation:

In your case:
`a=6`
`b=12`
`c=5`
`x_(1)=(-12+(12^2-4*6*5)^(1/2))/(2*6)=-0.59`
`x_2=(-12-(12^2-4*6*5)^(1/2))/(2*6)=-1.40`

Answer 2

`-0.5917` and `-1.408`

Explanation:

The x intercepts are basically the points where the line touches the x-axis. On the x-axis, the y co-ordinate is always zero so now we find values of x for which `6x^2 +12x + 5` = 0.

This is a quadratic equation and we can solve this using the quadratic formula:
`x` = `(-b+-sqrt(b^2-4*a*c))/(2*a)`

Now, for `6x^2+12x+5`,
a=6. b=12, c=5.
On substituting the values in the formula, we get

`x`= `(-12+-sqrt(12^2-4*6*5))/(2*6)`

`=` `(-12+-sqrt(144-120))/(12)`

`=` `(-12+-sqrt(24))/(12)`

This gives us the two values as `-0.5917` and `-1.408`

Hence the two `x` intercepts for the given equation are `-0.5917` and `-1.408`.