Examples with solutions for Difference of squares: Using roots

Exercise #1

Complete the following exercise:

(x+12)(x12)=0 (\sqrt{x}+\frac{1}{2})(\sqrt{x}-\frac{1}{2})=0

Video Solution

Step-by-Step Solution

To solve the equation (x+12)(x12)=0(\sqrt{x} + \frac{1}{2})(\sqrt{x} - \frac{1}{2}) = 0, we can apply the zero-product property, which tells us that if a product of two factors is zero, at least one of the factors must be zero.

Let us proceed with each factor:

  • First Factor: x+12=0\sqrt{x} + \frac{1}{2} = 0
    Solving for xx, subtract 12\frac{1}{2} from both sides:
    x=12\sqrt{x} = -\frac{1}{2}
    Squaring both sides, we get:
    x=(12)2=14x = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}.
    However, since the square root should be zero or positive, this case does not yield a real solution.
  • Second Factor: x12=0\sqrt{x} - \frac{1}{2} = 0
    Solving for xx, add 12\frac{1}{2} to both sides:
    x=12\sqrt{x} = \frac{1}{2}
    Squaring both sides, we have:
    x=(12)2=14x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.

Therefore, the solution to the equation (x+12)(x12)=0(\sqrt{x} + \frac{1}{2})(\sqrt{x} - \frac{1}{2}) = 0 is x=14x = \frac{1}{4}.

Upon reviewing the provided choices, the correct answer that matches our solution is: 14 \frac{1}{4} (Option 2).

Answer

14 \frac{1}{4}

Exercise #2

(x3)(x+3)(x+3)(x3)=5 \sqrt{(x-3)(x+3)}\cdot\sqrt{(x+3)(x-3)}=-5

Video Solution

Answer

2 2

Exercise #3

Resolve:

(7x)(x+7)=5x+423x47 (\sqrt{7}-\sqrt{x})(\sqrt{x}+\sqrt{7})=5x+42-3x-4\cdot7

Video Solution

Answer

1 1

Exercise #4

Resolve:

(x+3)(x3)+9x2=0 (\sqrt{x}+\sqrt{3})(\sqrt{x}-\sqrt{3})+\sqrt{9x^2}=0

Video Solution

Answer

32 \frac{3}{2}