What is the value of the real solution to the equation below? \sqrt{3x - 2} - \sqrt{x + 3}… — SAT

What is the value of the real solution to the equation below?

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

Answer: 6

The only real solution to the equation is 6.

Substituting

x = 6

into the original equation yields

\sqrt{3(6) - 2} - \sqrt{6 + 3} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1

, which is a true statement. Therefore, the only real solution is 6.

Isolate one of the radical terms

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

Preparing the equation to square both sides.

Square both sides of the equation

3x - 2 = x + 4 + 2\sqrt{x + 3}

Eliminating one of the square root radicals.

Isolate the remaining radical term and simplify

x - 3 = \sqrt{x + 3}

Simplifying the equation by isolating the second radical and dividing both sides by 2.

Square both sides again to eliminate the second radical

x^2 - 6x + 9 = x + 3

Converting the radical equation into a polynomial equation.

Write in standard quadratic form and factor

(x - 6)(x - 1) = 0

Setting the quadratic equation to zero and factoring to find potential solutions.

Verify solutions in the original equation to check for extraneous solutions

x = 6

is valid;

x = 1

is extraneous

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Solving equations containing radical expressions and identifying extraneous solutions.

Estimated Time:2m 30s

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