Solve the Radical Equations

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Exercise 1

\sqrt { 2x-3 } -x=-1

Exercise 2

\sqrt { 5x+4 } -1=2x

Exercise 3

3\sqrt { x-1 }+11=2x

Exercise 4

\sqrt { x } +\sqrt { x-4 } =2

Exercise 5

\sqrt { 2x-1 }+\sqrt { x+4 }=6

 

 

Solution of exercise 1

\sqrt { 2x-3 } -x=-1

\sqrt { 2x-3 }=-1+x

\sqrt { 2x-3 }=x-1

Taking square on both sides:

{ \sqrt { 2x-3 } }^{ 2 }={ (x-1) }^{ 2 }

2x-3={ (x) }^{ 2 }+2(x)(1)-{ (1) }^{ 2 }

2x-3=1-2x+{ x }^{ 2 }

1-2x+{ x }^{ 2 }-2x+3=0

1+3-2x-2x+{ x }^{ 2 }=0

4-4x+{ x }^{ 2 }=0

{ x }^{ 2 }-4x+4=0

You can solve the above equation with the help of middle term breaking or by using the quadratic formula. We will be using the quadratic formula in this question:

x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

a{ x }^{ 2 }+bx+c=0

{ x }^{ 2 }-4x+4=0

a=1,b=-4,c=4

 

x=\frac { -(-4)\pm \sqrt { { (-4) }^{ 2 }-4(1)(4) } }{ 2(1) }

x=\frac { 4\pm \sqrt { 16-16 } }{ 2 }

x=\frac { 4\pm 0 }{ 2 }

x=\frac { 4 + 0 }{ 2 } \qquad or \qquad x=\frac { 4 - 0 }{ 2 }

x=\frac { 4 }{ 2 } \qquad or \qquad x=\frac { 4 }{ 2 }

x=2

Solution of exercise 2

\sqrt { 5x+4 } -1=2x

\sqrt { 5x+4 } =2x+1

Taking square on both sides:

{ \sqrt { 5x+4 } }^{ 2 }={ (2x+1) }^{ 2 }

5x+4={ (2x) }^{ 2 }+2(2x)(1)+{ (1) }^{ 2 }

5x+4=4{ x }^{ 2 }+4x+1

4{ x }^{ 2 }+4x+1-5x-4=0

4{ x }^{ 2 }+4x-5x+1-4=0

4{ x }^{ 2 }-x-3=0

You can solve the above equation with the help of middle term breaking or by using the quadratic formula. We will be using the middle term breaking method in this question:

4{ x }^{ 2 }-x(4-3)-3=0

4{ x }^{ 2 }-4x+3x-3=0

4x(x-1)+3(x-1)=0

(4x+3)(x-1)=0

4x+3=0 \qquad or \qquad x-1=0

4x=-3 \qquad or \qquad x=1

x=\frac { -3 }{ 4 } \qquad or \qquad x=1

 

Substituting the values of x in the original equation to verify answers:

\sqrt { 5x+4 } -1=2x

When x=1:

\sqrt { 5(1)+4 } -1=2(1)

\sqrt { 5+4 } -1=2

\sqrt { 9 } -1=2

3-1=2

2=2

LHS=RHS, hence x=1 is a valid answer

 

When x=\frac { -3 }{ 4 }:

\sqrt { 5(\frac { -3 }{ 4 })+4 } -1=2(\frac { -3 }{ 4 })

\sqrt { \frac { -15 }{ 4 }+4 } -1=\frac { -3 }{ 2 }

\sqrt { \frac { -15 }{ 4 }+\frac { 16 }{ 4 } } -1=\frac { -3 }{ 2 }

\sqrt { \frac { -15+16 }{ 4 } } -1=\frac { -3 }{ 2 }

\sqrt { \frac { 1 }{ 4 } } -1=\frac { -3 }{ 2 }

\frac { 1 }{ 2 }-1=\frac { -3 }{ 2 }

\frac { 1 }{ 2 }-\frac { 2 }{ 2 }=\frac { -3 }{ 2 }

\frac { -1 }{ 2 }\neq \frac { -3 }{ 2 }

Hence, x=\frac { -3 }{ 4 } is not a valid answer.

Solution of exercise 3

3\sqrt { x-1 }+11=2x

3\sqrt { x-1 }=2x-11

{ 3\sqrt { x-1 } }^{ 2 }={ (2x-11) }^{ 2 }

9(x-1)={ (2x) }^{ 2 }-2(2x)(11)+{ (11) }^{ 2 }

9x-9=4{ x }^{ 2 }-44x+121

4{ x }^{ 2 }-44x+121-9x+9=0

4{ x }^{ 2 }-44x-9x+121+9=0

4{ x }^{ 2 }-53x+130=0

Using the quadratic formula to find the roots of the above equation:

x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

a{ x }^{ 2 }+bx+c=0

4{ x }^{ 2 }-53x+130=0

a=4,b=-53,c=130

 

x=\frac { -(-53)\pm \sqrt { { -53 }^{ 2 }-4(4)(130) } }{ 2(4) }

x=\frac { 53\pm \sqrt { 2809-2080 } }{ 8 }

x=\frac { 53\pm \sqrt { 729 } }{ 8 }

x=\frac { 53\pm 27 }{ 8 }

x=\frac { 53 + 27 }{ 8 } \qquad or \qquad x=\frac { 53 - 27 }{ 8 }

x=\frac { 80 }{ 8 } \qquad or \qquad x=\frac { 26 }{ 8 }

x=10 \qquad x=\frac { 26 }{ 8 }

x=10 \qquad x=\frac { 13 }{ 4 }

 

Verifying the above answers:

When x=10:

3\sqrt { x-1 }+11=2x

3\sqrt { 10-1 }+11=2(10)

3\sqrt { 9 }+11=20

3(3)+11=20

9+11=20

20=20

L.H.S=R.H.S

 

When x=\frac { -3 }{ 4 }:

3\sqrt { x-1 }+11=2x

3\sqrt { \frac { -3 }{ 4 }-1 }+11=2(\frac { -3 }{ 4 })

3\sqrt { \frac { -3 }{ 4 }-\frac { -4 }{ 4 } }+11=\frac { -6 }{ 4 }

3\sqrt { \frac { -3 }{ 4 }-\frac { -4 }{ 4 } }+11=\frac { -6 }{ 4 }

3\sqrt { \frac { -3-4 }{ 4 } }+11=\frac { -6 }{ 4 }

3\sqrt { \frac { -7 }{ 4 } }+11\neq \frac { -6 }{ 4 }

 

Hence, x=10 is the only valid answer.

Solution of exercise 4

\sqrt { x } +\sqrt { x-4 } =2

\sqrt { x }=2-\sqrt { x-4 }

{ \sqrt { x } }^{ 2 }={ 2-\sqrt { x-4 } }^{ 2 }

x={ (2) }^{ 2 }-2(2)(\sqrt { x-4 })+{ (\sqrt { x-4 }) }^{ 2 }

4-4(\sqrt { x-4 })+(x-4)-x=0

4-4(\sqrt { x-4 })+x-4-x=0

4-4-4(\sqrt { x-4 })+x-x=0

-4(\sqrt { x-4 })=0

\sqrt { x-4 }=0

{ \sqrt { x-4 } }^{ 2 }={ 0 }^{ 2 }

x-4=0

x=4

 

Verifying the answer:

\sqrt { x } +\sqrt { x-4 } =2

\sqrt { 4 } +\sqrt { 4-4 } =2

2+0=2

2=2

L.H.S=R.H.S

Solution of exercise 5

\sqrt { 2x-1 }+\sqrt { x+4 }=6

\sqrt { 2x-1 }=6-\sqrt { x+4 }

{ \sqrt { 2x-1 } }^{ 2 }={ 6-\sqrt { x+4 } }^{ 2 }

2x-1={ (6) }^{ 2 }-2(6)(\sqrt { x+4 })+{ (\sqrt { x+4 }) }^{ 2 }

2x-1=36-12\sqrt { x+4 }+x+4

2x-1=40-12\sqrt { x+4 }+x

2x-1-40-x=-12\sqrt { x+4 }

x-41=-12\sqrt { x+4 }

41-x=12\sqrt { x+4 }

{ 41-x }^{ 2 }={ 12\sqrt { x+4 } }^{ 2 }

{ (41) }^{ 2 }-2(41)(x)+{ (x) }^{ 2 }=144(x+4)

1681-82x+{ x }^{ 2 }=144x+576

{ x }^{ 2 }-82x+1681-144x+576=0

{ x }^{ 2 }-226x+1105=0

Using the breaking the middle term method to find the roots of the above equation:

{ x }^{ 2 }-x(221+5)+1105=0

{ x }^{ 2 }-221x-5x+1105=0

x(x-221)-5(x-221)=0

(x-5)(x-221)=0

x-5=0 \qquad x-221=0

x=5 \qquad x=221

 

Verifying both answers:

When x=5:

\sqrt { 2x-1 }+\sqrt { x+4 }=6

\sqrt { 2(5)-1 }+\sqrt { 5+4 }=6

\sqrt { 10-1 }+\sqrt { 5+4 }=6

\sqrt { 9 }+\sqrt { 9 }=6

3+3=6

6=6

R.H.S=L.H.S

 

When x=221:

\sqrt { 2x-1 }+\sqrt { x+4 }=6

\sqrt { 2(221)-1 }+\sqrt { 221+4 }=6

\sqrt { 442-1 }+\sqrt { 225 }=6

\sqrt { 441 }+\sqrt { 225 }=6

21+15=6

36\neq 6

Hence, x=5 is the only valid answer.

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.