Question

Please help? 2

Answer 1

See below

Explanation:

The quadratic formula is `x=(-b+-sqrtD)/(2a)`
Here `D = b^2 - 4ac`
Only to need to put the values in the formula.
a = 6
b = 5
c = -6

`x = [-5+-sqrt (5^2-4(6)(-6))]/(2*6)`
`x = [-5+-sqrt(25 + 144)]/12`
`x = [-5+-sqrt169]/12`
`x = [-5+-(13)]/12`
So x is either,
`(-5-13)/12`
=`-18/12`
=`-3/2`
Or
`(-5+13)/12`
=`8/12`
=`2/3`
Hope it helps you

Answer 2

See explanation.

Explanation:

1) `f(x)=6x^2+5x-6`
`=6x^2+9x-4x-6`
`=3x(2x+3)-2(2x+3)`
`=(2x+3)(3x-2)`

That's it for part1

2)
`f(x)=(-b+- sqrt(b^2-4ac))/(2a)`
Here, a=6, b=5, c=-6
Plugging in the values, the roots of the equation will be:
`(-5+- sqrt(5^2-4*6*(-6)))/(2*6`
Simplify the equation, and the roots will be
`(-5+- sqrt169)/12`
`=(-5+sqrt169)/12 or (-5-sqrt169)/12`
`=(-5+13)/12 or (-5-13)/12`
`=8/12 or -18/12`
`=2/3 or -3/2`

therefore, the equation will be:

`(x-2/3)(x+3/2)=0`

Thus, your final equation will be:
`(2x+3)(3x-2)`

`Thanks.`
Hope you got it.

Answer 3

Factoring Method

`color(blue)(f(x) = 6x^2+5x-6=(3x-2)(2x+3)`

Quadratic Formula

`color(blue)(x = 2/3, x= -3/2`

Explanation:

Given:

`color(green)(f(x) = 6x^2+5x-6`

The Standard Form of a Quadratic Equation:

`color(red)(y = f(x) = ax^2+bx+c = 0`

From our problem:

`a = 6; b = 5; and c = -6`

`color(brown)(Method.1)" "`Factoring Method

Using the Standard Form

`y = f(x) = ax^2+bx+c`

we find `color(blue)u` and `color(blue)v` such that

`color(green)(u *v = a*c and u + v = b`

Then we need to group them as shown below:

`ax^2 + ux + vx + c`

We have

`color(green)(f(x) = 6x^2+5x-6=0`

we find `color(blue)u` and `color(blue)v` as:

`color(green)(u = [-4] and v = [9]`

So, the middle term `color(blue)(5x)` can be written as `color(blue)([-4x+9x]`

We can now write our `f(x)` as

`color(green)(f(x) = 6x^2-4x+9x-6=0`

`rArr 6x^2-4x+9x-6=0`

`rArr 2x(3x-2)+3(3x-2)=0`

`rArr (3x-2)(2x+3)=0`

We get

`(3x-2) = 0, (2x+3) = 0`

`3x-2 rArr 3x = 2` hence `x= 2/3`

`2x+3 = 0 rArr 2x = -3` hence `x = -3/2`

Hence, `color(blue)(x=2/3, x = -3/2)`

`color(brown)(Method.2)" "`Using Quadratic Formula

Quadratic Formula is given by

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

From our problem:

`a = 6; b = 5; and c = -6`

Substituting these values of `a,b and c` in our formula

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

`rArr (-5+- sqrt(25+144))/12`

`rArr (-5+- sqrt(169))/12`

`rArr (-5+- 13)/12`

Hence,

`x = (-5+13)/12, x = (-5-13)/12`

`x = 8/12, x = -18/12`

`x = 2/3, x = -3/2`

Hence, `color(blue)(x=2/3, x = -3/2)`

We can observe that both the methods yield the same values for `x`

Hope you find this solution helpful.