Given: \( x + 5 = \sqrt{x + 35} \) Step 1: Square both sides to eliminate the square root. \( (x + 5)^2 = (\sqrt{x + 35})^2 \) Step 2: Expand the left side and simplify the right side. \( x^2 + 10x + 25 = x + 35 \) Step 3: Set the equation to zero and rearrange terms. \( x^2 + 10x + 25 - x - 35 = 0 \) \( x^2 + 9x - 10 = 0 \) Step 4: Factor the quadratic equation. \( (x + 10)(x - 1) = 0 \) Step 5: Solve for x. \( x + 10 = 0 \) or \( x - 1 = 0 \) \( x = -10 \) or \( x = 1 \) Step 6: Check for extraneous solutions by plugging back into the original equation. For \( x = -10 \): \( -10 + 5 \neq \sqrt{-10 + 35} \) \( -5 \neq 5 \) (Not a solution) For \( x = 1 \): \( 1 + 5 = \sqrt{1 + 35} \) \( 6 = \sqrt{36} \) \( 6 = 6 \) (Valid solution) Therefore, \( x = 1 \) is the solution.
To solve the equation involving the radical, we want to isolate p. Given the equation from the image: \[ \sqrt{\frac{2p+1}{t}} = 3 \] First, square both sides of the equation to eliminate the square root: \[ \left( \sqrt{\frac{2p+1}{t}} \right)^2 = 3^2 \] This simplifies to: \[ \frac{2p+1}{t} = 9 \] Next, multiply both sides by t to remove the denominator: \[ 2p+1 = 9t \] Subtract 1 from both sides to isolate the term with p: \[ 2p = 9t - 1 \] Finally, divide both sides by 2 to solve for p: \[ p = \frac{9t - 1}{2} \] Therefore, the value of p in terms of t is: \[ p = \frac{9t - 1}{2} \]
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