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sqrt(2x-3)-sqrt(x+7)+2=0, solve for x











































sqrt(2x - 3) - sqrt(x + 7) + 2 = 0

this particular equation can be rewritten in exactly the same format as a quadratic equation.

ax² + bx +c = 0

When in this format, it can be easily factored:

(     )(     ) = 0

There are two solutions, but only one solution is valid.

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Aug 26, 2011
Calculus - Equation with Radicals
by: Staff

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Part II


Factor the equation

(x - 2)(x - 42) = 0



1st solution:

“Divide” each side of the equation “by (x - 42)”, and then solve for x

(x - 2)(x - 42)/(x - 42) = 0/(x - 42)

(x - 2)*[(x - 42)/(x - 42)] = 0/(x - 42)

(x - 2)*1 = 0/(x - 42)

(x - 2) = 0/(x - 42)

x - 2 = 0/(x - 42)

x - 2 = 0

x - 2 + 2 = 0 + 2

x + 0 = 0 + 2

x = 0 + 2

x = 2



2nd solution:

“Divide” each side of the equation “by (x - 2)”, and then solve for x

(x - 2)(x - 42)/(x - 2) = 0/(x - 2)

(x - 42)*[(x - 2)/(x - 2)] = 0/(x - 2)

(x - 42)*1 = 0/(x - 2)

(x - 42) = 0/(x - 2)

x - 42 = 0/(x - 2)

x - 42 = 0

x - 42 + 42 = 0 + 42

x + 0 = 0 + 42

x = 0 + 42

x = 42→ x = 42 is a solution, but NOT a VALID SOLUTION (read on)



The final answer is: x = 2



Check this equation graphically. A graphical solution clearly shows that x = 2 is the only valid solution.


(1) Click the following link to VIEW the graphical solution; or (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

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http://www.solving-math-problems.com/images/rad-eq-2011-08-26-a.png


check both solutions by substituting the numerical values of x into the original equation


for x = 2

√(2x-3) - √(x+7) + 2 = 0

√(2*2-3) - √(2+7) + 2 = 0

√(4-3) - √(2+7) + 2 = 0

√(1) - √(2+7) + 2 = 0

1 - √(2+7) + 2 = 0

1 - √(9) + 2 = 0

1 - 3 + 2 = 0

0 = 0, OK → x = 2 is a valid solution


for x = 42

√(2x-3) - √(x+7) + 2 = 0

√(2*42-3) - √(42+7) + 2 = 0

√(84-3) - √(42+7) + 2 = 0

√(81) - √(42+7) + 2 = 0

9 - √(42+7) + 2 = 0

9 - √(49) + 2 = 0

9 - 7 + 2 = 0

4 ≠ 0, NO → x = 42 is NOT a VALID SOLUTION





Thanks for writing.

Staff
www.solving-math-problems.com



Aug 26, 2011
Calculus - Equation with Radicals
by: Staff


Part I

The question:

sqrt(2x-3)-sqrt(x+7)+2=0


The answer:

sqrt(2x-3) - sqrt(x+7) + 2 = 0

√(2x-3) - √(x+7) + 2 = 0

Subtract 2 from each side of the equation

√(2x-3) - √(x+7) + 2 - 2 = 0 - 2

√(2x-3) - √(x+7) + 0 = 0 – 2

√(2x-3) - √(x+7) = 0 - 2

√(2x-3) - √(x+7) = - 2

Add √(x+7) to each side of the equation

√(2x-3) - √(x+7) + √(x+7) = - 2 + √(x+7)

√(2x-3) + 0 = - 2 + √(x+7)

√(2x-3) = - 2 + √(x+7)

√(2x-3) = √(x+7) - 2

Square both sides of the equation

[√(2x-3)]² = [√(x+7) - 2]²

2x - 3 = x + 7 - 4√(x+7) + 4

Subtract x from each side of the equation

2x - 3 - x = x + 7 - 4√(x+7) + 4 - x

2x - x - 3 = x + 7 - 4√(x+7) + 4 - x

x - 3 = x + 7 - 4√(x+7) + 4 - x

x - 3 = x - x + 7 - 4√(x+7) + 4

x - 3 = 0 + 7 - 4√(x+7) + 4

x - 3 = 7 - 4√(x+7) + 4

x - 3 = 7 + 4 - 4√(x+7)

x - 3 = 11 - 4√(x+7)

subtract 11 from each side of the equation

x - 3 - 11 = 11 - 11 - 4√(x+7)

x - 14 = 0 - 4√(x+7)

x - 14 = - 4√(x+7)

Square each side of the equation

(x - 14)² = [- 4√(x+7)]²

(x - 14)² = (- 4)² * [√(x+7)]²

x² - 28x + 196 = 16 * (x + 7)

x² - 28x + 196 = 16x + 112


Subtract 16x from each side of the equation

x² - 28x + 196 - 16x = 16x + 112 - 16x

x² - 28x - 16x + 196 = 16x + 112 - 16x

x² - 44x + 196 = 16x + 112 - 16x

x² - 44x + 196 = 16x - 16x + 112

x² - 44x + 196 = 0 + 112

x² - 44x + 196 = 112


Subtract 112 from each side of the equation

x² - 44x + 196 - 112 = 112 - 112

x² - 44x + 84 = 112 - 112

x² - 44x + 84 = 0

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