Please help? 2

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3 Answers
Jan 18, 2018

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

Jan 18, 2018

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.

Jan 18, 2018

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.