a. 2x = sqr root 12x+72 b. sqr root x+5 = 5 - sqr root x c. |2x| = -|x+6| d. 5 = |x+4| + |x-1| e. |x-2| = 4 - |x-3| 1. Solve the following equations: a. 2x = sqr root 12x+72...

a. 2x = sqr root 12x+72
b. sqr root x+5 = 5 - sqr root x
c. |2x| = -|x+6|
d. 5 = |x+4| + |x-1|
e. |x-2| = 4 - |x-3|

1. Solve the following equations:

a. 2x = sqr root 12x+72
b. sqr root x+5 = 5 - sqr root x
c. |2x| = -|x+6|
d. 5 = |x+4| + |x-1|
e. |x-2| = 4 - |x-3|

 

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neela | High School Teacher | (Level 3) Valedictorian

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a. 2x = sqr root 12x+72 Solution: 2x=sqrt12x+72. Or
2x-72 =sqrt12 squaring both sides, (2x-72)^2 = 12x^2. 4x^2-288x+72^2 = 12x. Or Dividing by 4, x^2-72x+36^2= 3x. Or x^2-75x+36^2 = 0, x1 = [75+sqrt(75^2-4*36^2)]/2 =(75+21)/2 = 48 Or x2=(75-21)/2 = 54/2 = 27. x1 =48 is valid root and x2 = 27 is an extraneous root.

b. sqr root x+5 = 5 - sqr root x sqrtx+5 = 5-sqrtx. Colecting sqrtx together, 2sqrtx = 0. So, sqrt x = 0. Therefore x = 0.

c. |2x| = -|x+6| Obviously there is no solution as LHS is positive and RHS is -ve, unless both sides are zero which does not happen for the same value of x on LHS and RHS. Detatailed case wise analysis is depicted below:
If x>0,
2x =-(x+6). Or 3x= -6. Or x = -6/3 = -2 is a contradiction, as x>0 and x =-2 does not hold. If x=0, the equation does not hold good. if -6=<x<0 -2x= -(x+6).Or -2x+x= -6. Or -x = -6 Or x=3 and -6=<x<0  are inconsistent. If x<-6, -2x= -(6-x). Or -2x-x = -6. Or -3x=-6. Or x =2 is also contradiction to x<-6. So no solution.

d. 5 = |x+4| + |x-1| Intuitively x= 0 is a solution. If x>=1, 5 = x+4+x-1. 5-3 = 2x Or x =1 is a valid solution. When 0<=x<=1, So, 0<=x<=1 is solution. Any value in the closed interval (0,1) is a solution.
5=x+4+1-x =5 is an identity. If -4<=x<0. 5=x+4+1-x =5 is an identity. So for any value for which 4<=x<0,  is a a solution. If x <-4, 5 = 4-x+1-x. Or 0< -2x or x=0 is not intune with x<-4. But x=0 is already covered earlier. So -4<=x<0  or 0<=x<=1 are solutions. Or combining the intervals, we get -4<=x<=1 is the solution.

e. |x-2| = 4 - |x-3| If x>3, x-2 = 4-(x-3) Or 2x =4+3+2. Or x =4.5 is a solution. If 2<x<3, x-2= 4-(3-x). Or x-2 = 1+x, Or -2 = 1 a contradicting situation. So 2<x<3 cannot be a solution.
If 0<x<2, 2-x = 4-(3-x). Or 2-x = 1+x. Or 2-1 = 2x Or x=1/2 is a solution , consistent with 0<x<2. If x<0, 2-x = 4-(3-x) . Or 2-1 = 2x Or x =1/2 is inconsistent under x<0 case. But x= 1/2 is already covered under different case where it was valid.
So summerising , x=4.5 or x=1/2 are the valid solutions.

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