How do you solve #sqrt( 9x+81)= x+5#?

Answer 1

7 and -8

Square both sides --> #9x + 81 = (x + 5)^2 = x^2 + 10 x + 25.# #x^2 + x - 56 = 0.# Solve this quadratic equation by using the new Transforming Method (Socratic Search). Find 2 numbers knowing sum (-1) and product (-56). The roots have opposite signs because ac < 0. Factor pairs of (-56) --> ...(-7, 8)(7, -8). This last sum is -1 = -b. Then, the 2 real roots are: 7 and -8.

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Answer 2

Savannah Anderson

To solve the equation sqrt(9x+81) = x+5, we can follow these steps:

Square both sides of the equation to eliminate the square root: (sqrt(9x+81))^2 = (x+5)^2.
Simplify the equation: 9x+81 = x^2 + 10x + 25.
Rearrange the equation to form a quadratic equation: x^2 + x - 56 = 0.
Factorize the quadratic equation: (x+8)(x-7) = 0.
Set each factor equal to zero and solve for x: x+8 = 0 or x-7 = 0.
Solve for x in each equation: x = -8 or x = 7.

Therefore, the solutions to the equation sqrt(9x+81) = x+5 are x = -8 and x = 7.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.