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 clear-up the forthcoming gists:

1.      Clear-up for x 




To clear-up for x in aloft gist we conquer proportion the genuine roots for the aloft quadratic equations by looking for two bulk whose sum is 1 and result is 12, then we equate them in the ancient equation and clear-up for x.

Sum = 1, result = 12 the two bulk are 4 and -3

We equate to bear 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.                  Clear-up for x    22x-4=64


Too clear-up for x we primeval equate the two esteems on twain sides to the selfselfidentical deep (deep 2) then past the two are correspondent  and after a while selfselfidentical deep then their powers are correspondent thus we equate the powers and clear-up

Solve for x    22x-4=64    hush 64=26

                   22x-4=26 thus 2x-4=6, gather affect conditions together

                              2x= 6+4

2x=10 and diving by 2 twain sides

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


3.                  Clear-up for x

3cosx +2sin2x=0


3cosx +2sin2x=0  

To clear-up for x we conquer use the trigonometric imparity stating that cos2x + sin2x =1 thus sin2x=1-cos2x thus we re-establish sin2x after a while

 1-cos2x to bear

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

3y-2y2+2=0, -2y2=3y+2=0 then we clear-up the quadratic equation

Sum 3, result= -4 the two bulk 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 esteem of x is120, x=120 or x = 240