What is x^2+( 5√2-x)^2= 5√2^2

1 answer:

3 0

$x^2+( 5\sqrt2-x)^2= (5\sqrt2)^2\\
x^2+50-10\sqrt2x+x^2=50\\
2x^2-10\sqrt2x=0\\
x^2-5\sqrt2x=0\\
x(x-5\sqrt2)=0\\
x=0 \vee x=5\sqrt2$

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The complete question is:

A ball is thrown from an initial height of 2 feet with an initial upward velocity of 31 ft/s. The ball's height h (in feet) after 7 seconds is given by the following, h=2+31t-16t². Find all values of t for which the ball's height is 16 feet. Round your answer(s) to the nearest hundredth.