Answer:
[tex]d=-18[/tex]
Step-by-step explanation:
The only way we can achieve an extraneous solution is by squaring both sides. Example:
[tex]\sqrt{-1}=x, \\\sqrt{-1}^2=x^2,\\1=x^2,\\x=1\text{ [extraneous]}[/tex]
Square both sides of the equation:
[tex]\sqrt{\frac{1}{2}y-1}^2=(\frac{3}{4}y+d)^2[/tex]
Substitute [tex]y=20[/tex]:
[tex]9=(15+d)^2[/tex]
Expand the right side using [tex](a+b)^2=a^2+2ab+b^2[/tex]:
[tex]9=15^2+2(15)(x)+x^2,\\x^2+30x+225=9[/tex]
Subtract 9 from both sides:
[tex]x^2+30x+216=0[/tex]
Factor:
[tex](x+12)(x+18)=0,\\\begin{cases}x+12=0, x=\boxed{-12},\\x+18=0,x=\boxed{-18}\end{cases}[/tex]
Substitute both solutions to see which work:
[tex]\sqrt{\frac{1}{2}(20)-1}=(\frac{3}{4}(20)+d), \\\\d=-12\checkmark\\d=-18\times[/tex]
The solution [tex]d=-18[/tex] yields [tex]3=-3[/tex] which does not work and therefore is extraneous.
