## Algebra 1

Given Infomation : $\sqrt 2r - 3 = r$ ------- eq1 By squaring the eq 1 we get $2r- 3= r^{2}$ ---- eq 2 By simplifying the eq 2 on both sides we get, $0=r^{2}-2r+3$ ----eq3 Now, by factoring the eq 3 $0=(r-3)(r+1)$ By using the zero product property $r-3=$ -> r=3 $r+1$ -> r=-1 Now, By checking At r=3, $\sqrt 2(3)-3=3$ $\sqrt 3\ne3$ r=-1 $\sqrt 2(-1)-3=-1$ $\sqrt -5\ne-1$ So Both are extraneous solutions. Hence there are no solutions