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.