Answer:
d=2
Explanation:
We are given equation:
sqrt(4y-3)=d-y
Squaring both sides gives:
4y-3=(d-y)^2
Applying the identity (x+y)^2=x^2+2xy+y^2 on right:
4y-3=d^2-2dy+y^2
Now let's y=7:
4(7)-3=d^2-2d(7)+(7)^2
Simplify:
25=d^2-14d+49
Subtract 25 on both sides:
0=d^2-14d+24
Factor left:
0=(d-12)(d-2) since -2+-12=-14 and -2(-12)=24
This gives d=12 or d=2.
The d that makes d-y or I mean d-7 negative will give us y=7 as extraneous
Since 2-7 is -5, then d=2 is what we are looking for.
Check:
sqrt(4y-3)=d-y
Set d=2: sqrt(4y-3)=2-y
Now solve for y:
Square both sides: 4y-3=4-4y+y^2
Subtract 4y and 3 on both sides: 0=7-8y+y^2
Reorder right side: 0=y^2-8y+7
Factor: 0=(y-7)(y-1) since -7+-1=-8 and -7(-1)=7
This gives y=7 or y=1.
Plugging in y=7 for a check to this equation gives:
sqrt(4×7-3)=2-7
Sqrt(25)=-5
5=-5 which is not true which is what we wanted