Question

Can I get some help finding the x - intercept please? Thanks!

Answer 1

`x_1=-1+1/sqrt(6)` and `x_2=-1-1/sqrt(6)`

Explanation:

`6x^2+12x+5=0`

This can be solved using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

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

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

`x=-1+-sqrt(24)/12=-1+-sqrt(6times4)/12`

`x = -1+(2sqrt(6))/12=-1+-sqrt(6)/6`

`x=-1+-1/sqrt(6)`

So the `x` intercepts are:

`x_1=-1+1/sqrt(6)` and `x_2=-1-1/sqrt(6)`

Answer 2

`(-6+sqrt6)/6` and `(-6-sqrt6)/6`, or `-0.59` and `-1.41`

Explanation:

Use the quadratic formula to find the x-values of the equation.

`x=(-(12)+-sqrt((12^2)-4(6)(5)))/(2(6)`
Simplify the radicand to get `sqrt(144-120)`, which simplifies to `sqrt24`, or `2sqrt6`.
Simplify the rest of the equation to get `(-12+-2sqrt6)/12` which further simplifies to `(-6+-sqrt6)/6`.

Make sure you understand why you are doing this!

You are basically making the function equal to zero because whenever the function crosses the x axis, the x value is 0. Therefore, you are trying to find for what values the x equals zero.

Answer 3

This is effectively asking us to solve `6x^2+12x+5=0`

The first way of trying to solve a quadratic equation is to factorise.

We need two numbers that sum to `12` and multiply to `5xx6=30`. Try thinking of some.

In fact, there are no such numbers. So next, we turn to our two other ways of solving quadratic equations.

Method 1 - Completing the Square

This method is good if you don't have a calculator, and ideal if the coefficient of `x^2` is 1 or a factor of the `x` coefficient.

`6x^2+12x+5=0`
`x^2+2x+5/6=0`
`(x+1)^2-1+5/6=0`
`(x+1)^2-1/6=0`
`(x+1)^2=1/6`
`x+1=+-sqrt(1/6)`
`x+1=+-1/sqrt6`
`x+1=+-sqrt6/6`
`x=-1+-sqrt6/6`
Tidy up
`x=(-6+-sqrt6)/6`

So `x=(-6+sqrt6)/6` and `x=(-6-sqrt6)/6`

Method 2 - The Quadratic Formula

A much better method if you have a calculator. Given the choice, the formula is a much better option.

This states that, for a quadratic `ax^2+bx+c=0`, the solutions are given by:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

So for this example, `a=6, b=12` and `c=5`

`x=(-12+-sqrt(12^2-4xx6xx5))/(2xx6)`
`x=(-12+-sqrt(144-120))/12`
`x=(-12+-sqrt24)/12`
`x=(-cancel12color(red)(6)+-cancel2sqrt6)/(cancel(12)color(red)(6))`
`x=(-6+-sqrt6)/6`

So which method is better to use? Factorising is the best method, although it is not always possible to factorise. Personally, the quadratic formula is the better method if you cannot factorise, since there are fewer steps involved compared to completing the square.

Can I factorise?

Look at the quadratic formula again.

`x=(-b+-sqrt(color(red)(b^2-4ac)))/(2a)`

The area under the square root sign can tell us about the nature of the solutions. This expression, `b^2-4ac`, is known as the discriminant. The symbol `Delta` (capital delta) may be used to refer to this.

-if `Delta>0`, then there are two distinct real roots (2 solutions)
-if `Delta=0` then there is one repeated real root (1 solution)
-if `Delta<0` then there are no real roots (does not cross the x axis). This is because we are square rooting a negative, which can be a no-go.

If a quadratic factorises, then `b^2-4ac` will be a positive square number since when we square root it, there will be no surd. In our example:

`b^2-4ac=12^2-4xx6xx5=24`
When we square root this, we get `sqrt24=2sqrt6`. Since we have a surd, this won't factorise. This is why there are no numbers to make such an expression in the beginning.