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Please solve the following problems:

- Solve for x

X^{2}+x-12=0

__SOLUTION__

X^{2}+x-12=0

To solve for x in above problem we will calculate the real roots for the above quadratic equations by looking for two numbers whose sum is 1 and product is 12, then we equate them in the original equation and solve for x.

Sum = 1, product = 12 the two numbers are 4 and -3

We equate to have x^{2}+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

- Solve for x 2
^{2x-4}=64

__SOLUTION__

Too solve for x we first equate the two values on both sides to the same base (base 2) then since the two are equal and with same base then their powers are equal thus we equate the powers and solve

Solve for x 2^{2x-4}=64 note 64=2^{6}

2^{2x-4}=2^{6 }thus 2x-4=6^{, }collect like terms together

2x= 6+4

2x=10 and diving by 2 both sides

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

X=5

- Solve for x

3cosx +2sin^{2}x=0

__Solution__

3cosx +2sin^{2}x=0

To solve for x we will use the trigonometric inequality stating that cos^{2}x + sin^{2}x =1 thus sin^{2}x=1-cos^{2}x thus we replace sin^{2}x with

1-cos^{2}x to have

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

3y-2y^{2}+2=0, -2y^{2}=3y+2=0 then we solve the quadratic equation

Sum 3, product= -4 the two numbers are 4 and -1

-2y^{2}+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 recall 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 value of x is120, x=120 or x = 240

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