Question

Solving Rational and Radical Equations: Mastery Test

Answer 1

Answer:

1 step: Raise both sides of the equation to the power of 2.

2 step: Simplify to obtain the final radical term on one side of the equation.

3 step: Raise both sides of the equation to the power of 2 again.

4 step: Simplify to get a quadratic equation.

5 step: Use the quadratic formula to find the values of x.

6 step: Apply the Zero Product Rule.

Step-by-step explanation:

Given the equation

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

1 step: Raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2\\ \\(\sqrt{x+3})^2-2\sqrt{x+3}\sqrt{2x-1}+(\sqrt{2x-1})^2=4\\ \\x+3-2\sqrt{x+3}\sqrt{2x-1}+2x-1=4$

2 step: Simplify to obtain the final radical term on one side of the equation.

$x+3-2\sqrt{x+3}\sqrt{2x-1}+2x-1=4\\ \\3x+2-2\sqrt{x+3}\sqrt{2x-1}=4\\ \\-2\sqrt{x+3}\sqrt{2x-1}=4-3x-2\\ \\-2\sqrt{x+3}\sqrt{2x-1}=2-3x$

3 step: Raise both sides of the equation to the power of 2 again.

$(-2\sqrt{x+3}\sqrt{2x-1})^2=(2-3x)^2\\ \\4(x+3)(2x-1)=(2-3x)^2$

4 step: Simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=2^2-2\cdot 2\cdot 3x+(3x)^2\\ \\8x^2-4x+24x-12=4-12x+9x^2\\ \\8x^2+20x-12-4+12x-9x^2=0\\ \\-x^2+32x-16=0\\ \\x^2-32x+16=0$

5 step: Use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 1\cdot 16=1,024-64=960\\ \\x_{1,2}=\dfrac{-(-32)\pm \sqrt{960}}{2\cdot 1}=\dfrac{32\pm 8\sqrt{15}}{2}=16\pm 4\sqrt{15}$

Then the equation is

$(x-16-4\sqrt{15})(x-16+4\sqrt{15})=0$

6 step: Apply the Zero Product Rule.

$(x-16-4\sqrt{15})(x-16+4\sqrt{15})=0\\ \\x-16-4\sqrt{15}=0\text{ or }x-16+4\sqrt{15}=0\\ \\x_1=16+4\sqrt{15}\text{ or }x_2=16-4\sqrt{15}$