Question

sert y=0 and solve 4x- (12-4x)^1/2

Answer 1

y = 4x - (12-4x)^1/2

if y = 0

\[0= 4x - \sqrt{12-4x}\]
\[\sqrt{12-4x} = 4x\]

square both sides

\[12-4x = 16 x ^{2}\]
\[12 = 16x ^{2} + 4x\]

then divide by 16

\[\frac{ 3 }{ 4 } = x ^{2} +\frac{ 1 }{ 4 }x\]

completing the squares

\[\frac{ 3 }{ 4 } + \frac{ 1 }{ 64 } = (x + \frac{ 1 }{ 8 })^{2}\]

\[\sqrt{\frac{ 49 }{ 64 }} = x + \frac{ 1 }{ 8 }\]

\[-\frac{ 1 }{ 8 } \pm \frac{ 7 }{ 8 } = x\]

x can be

x= -1
x = 3/4

Answer 2

correct me if im wrong to something sorry i have no papers here and im on a shop

Answer 3

yup x=-1 or x=3/4. But x=3/4 is the only answer where
4(3/4) - (12-4(3/4))^1/2 = 0 which satisfies with your y = 0

But when x=-1
4(-1) - (12-4(-1))^1/2 = -8 where it fails -8 is not equal to 0..

So x=3/4 is the only answer