Question 1075997: A 48-foot ladder resting against a house reaches a windowsill 32 feet above the ground. How far is the foot of the ladder from the foundation of the house? Write answer in simplest radical form. [Hint: Draw the diagram and us Pythagoras’ theorem] [8mks]
Solve: 4√(x+1)-2√(x+6)=2 [10mks]
Rationalize and simplify:
(3√y)/(-3√y+2√x)
[8mks]
Simplify the following complex fraction:
(1+n/(n-2))/(3/(n^2-4)+2)
[6mks]
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The hypotenuse is the ladder, 48', and the vertical leg is 32 feet. The horizontal leg needs to be determined. That is what should be drawn.
square of 48=square of 32 + square of x
2304=1024+ square of x
1280=square of x
x=sqrt (256)*sqrt (5)=16 sqrt (5)'
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4√(x+1)-2√(x+6)=2
4 sqrt (x+1)=2+2 sqrt (x+6). Isolate each radical. Now square both sides
16(x+1)=4+8 sqrt(x+6)+4*(x+6)=4+8 sqrt (x+6)+4x+24
16x+16=28+4x+8 sqrt(x+6)
12x-12=8 sqrt(x+6), isolating the radical
divide by 3
3x-3=2 sqrt (x+6)
9x^2-18x+9=4(x+6), squaring both sides
9x^2-22x-15=0
(9x+5)(x-3)=0
x=-5/9, 3
x=3; 4*2-2*3=2, which checks
x=-5/9; 4 sqrt (4/9)-2*sqrt(49/9)=2; 4*(2/3)-2(7/3)=8/3-14/3=-2 doesn't check
x=3
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3 sqrt (y) divided by
-3 sqrt (y)+ 2 sqrt (x)
multiply top and bottom by the denominator conjugate, 3 sqrt (y)+2 sqrt (x)
the numerator is 9y+6 sqrt (xy)
the denominator is 4x-9y
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