Algebra Solution.docx - Please solve the following problems 1 Solve for x x x-12=0 2 Solution x={−4,3 a We need to factorize this expression finding 2

Algebra Solution.docx - Please solve the following problems...

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Please solve the following problems: 1. Solve for x x 2 +x-12=0 Solution: x ={−4,3} a) We need to factorize this expression finding 2 numbers such that: (x+N1) (x+N2)= x2+x-12, where N1+N2=1 and (N1*N2)=-12 Those numbers are: N1=4 and N2=-3 because 4+( -3)=1 and 4 x (-3)= -12 Therefore, the factorized expression is (x+4)(x−3) b) Now we equate this factorized expression to zero (x+4)(x−3)= 0 if (x+4)=0 ; x= -4 if (x-3)=0 ; x=3 2. Solve for x 2 2x-4 =64 Solution: x=5 a) We need to find an exponent with the same base, which gives 64 2 n =64 This number is 6, because 2 6= 64= 2x2x2x2x2x2 Therefore, 2 2x-4 =2 6 b) Since the bases are the same, the two expressions are only equal if the exponents are also equal 2x-4=6 c) Now we can solve the equation for x. First, we add 4 in both sides 2x-4+4=6+4 2x=10 Then, we divide by 2 in both sides 2x/2= 10/2
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x=5 3. Solve for x 3 cos (x) +2sin 2 (x)=0 Solution: x= 2 π 3 or 120° a) We know the identity sin
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Unformatted text preview: 2 x+cos 2 x=1, we can write it as sin 2 x =1-cos 2 x b) Substitute this in our equation 3 cos (x) +2sin 2 (x)=0; 3 cos x +2( 1-cos 2 x )= 0 3 cos x +2 -2cos 2 x = 0 We can write it as -2cos 2 x+3 cos x +2=0 If we suppose cos(x)=y; we can write it as-2y 2 +3y+2=0 c) This expression follows the form ax 2 +bx+c=0, which can be solved using the quadratic formula ¿ − b± √ b 2 − 4 ac 2 a , where a=-2, b=3 and c=2 y = − 3 ± √ 3 2 − 4 ( − 2 ) 2 2 (− 2 ) y = − 3 ± √ 25 − 4 y=-1/2 and y=2 Hence, cos (x)= -1/2 or cos (x)= 2 d) First, we solve cos (x)= -1/2. We do the inverse of the cos of both sides of the equation to extract x inside cos. The inverse of cos (x) is arccos(x) X=arccos (-1/2) X= 2 π 3 = 120° Second, we must solve cos (x)= 2. But there is no solution for this x in R....
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