Please solve the following problems: 1. Solve for x X2+x-12=0 SOLUTION X2+x-12=0 To solve for x in above problem we will calculate the real roots for the above quadratic equations by looking for


Please work-out the subjoined problems:

1.      Work-out for x 




To work-out for x in over problem we achieve weigh the legitimate roots for the over quadratic equations by looking for two total whose sum is 1 and issue is 12, then we equate them in the initiatory equation and work-out for x.

Sum = 1, issue = 12 the two total are 4 and -3

We equate to entertain x2+4x-3x-12=0

                            X(x+4)-3(x+4) = 0 thus x+4=0   and X= -4

OR      X-3=0 AND X=3 thus x = 3 or -4

2.                  Work-out for x    22x-4=64


Too work-out for x we foremost equate the two computes on twain sides to the selfselfidentical disesteemed (disesteemed 2) then since the two are resembling  and delay selfselfidentical disesteemed then their powers are resembling thus we equate the powers and work-out

Solve for x    22x-4=64    still n ess 64=26

                   22x-4=26 thus 2x-4=6, congregate enjoy conditions together

                              2x= 6+4

2x=10 and diving by 2 twain sides

2x/2 = 10/2, x=5


3.                  Work-out for x

3cosx +2sin2x=0


3cosx +2sin2x=0  

To work-out for x we achieve use the trigonometric disparity stating that cos2x + sin2x =1 thus sin2x=1-cos2x thus we restore sin2x delay

 1-cos2x to entertain

3cosx+2(1-cos2x) =0, 3cosx+2-2cos2x = 0, let cos x be y

3y-2y2+2=0, -2y2=3y+2=0 then we work-out the quadratic equation

Sum 3, issue= -4 the two total are 4 and -1

-2y2+4y-y+2=0 , -2y(y-2)-1(y-2) =0  ,(-2y-1)(y-2)=0

We equate in (-2y-1)=0 ,y=-1/2 or y-2=0  ,y=2

So -2y = -1, y= -1/2   or y-2=0 , y=2 foreclosure y= cosx  thus cosx=2

Recall y= cos (x) thus y=cos x = -1/2  or 2

x=cos-1 2 which is absurd

X= cos-1 -1/2 = 120 or 240

Thus compute of x is120, x=120 or x = 240

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