9514 1404 393
Answer:
(c) Both equations have the same potential solutions, but equation A might have extraneous solutions.
Step-by-step explanation:
The general approach to solving an equation like either of these is to raise both sides of the equation to a power that will remove the radical. In both cases, the result is a quadratic with roots of x=-4 and x=2.
Of these two potential solutions, x = -4 is an extraneous solution for equation A. Both values of x are solutions for equation B. An appropriate description is ...
Both equations have the same potential solutions, but equation A might have extraneous solutions.
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<em>Additional comment</em>
The attached graph shows the equations cast into the form f(x) = 0, so x-intercepts are the solutions to the equation. The radical versions of the equations have only x-intercepts that are actual solutions. The version with the radicals removed is a parabola with two solutions (orange). Only one of those matches the solution to equation A (red). Both match the solutions of equation B (purple).
Answer:c
Step-by-step explanation:
Answer:
y = √(3² + 3²)
9 = √(5² + x²)
Step-by-step explanation:
Diagram 1:
Hypotenuse = y
Opposite = 3
Adjacent = 3
y² = 3² + 3²
y = √(3² + 3²)
y = √(9 + 9)
y = √18
Diagram 2:
Hypotenuse ² = known leg² + unknown leg²
Hypotenuse = 9
Known leg = 5
Unknown leg = x
9² = 5² + x²
9 = √(5² + x²)
9² = 5² + x²
9² - 5² = x²
x² = 81 - 25
x = √ 56
S = 7x^2....divide both sides by 7
1/7s (same as s/7) = x^2...now take sqrt of both sides
(+ - ) sqrt 1/7s = x