If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
\(0 < 4x^2 < 4\)

Lets check the options now,
(A) 1 (\(0 < 4x^2 < 4\))
(B) 4 (\(0 < 4x^2 < 4\))
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but \(0 < 4x^2 < 4\))
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but \(0 < 4x^2 < 4\))
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and \(0 < 4x^2 < 4\))

Thus, only possible option which lies in the range is E.

Answer (E),

Regards,

Hi narangvaibhav,

Yes definitely, we can use the conventional method too,
\(\sqrt{3-2x} = \sqrt{2x} +1\)
Squaring both the sides,
\(3-2x=2x+1+2 \sqrt{2x}\)
or \(1-2x=\sqrt{2x}\)
Again, squaring both the sides,
\(1+4x^2-4x=2x\)
or \(4x^2=6x-1\)

Answer (E).

Regards,

PS: But that isn't much fun.

Can someone help me solve this?

\(\sqrt{(3-2x)}\) = \(\sqrt{2x}\) + 1
\(\sqrt{(3-2x)}\) + 1 = \(\sqrt{2x}\)
If I square both sides,
2 - \(\sqrt{(3-2x)}\) = 2x
So, \(4x^2\) = 4 - (3-2x) - 2\(\sqrt{(3-2x)}\)

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

Please help

You could solve this way too:

\(\sqrt{3-2x} = \sqrt{2x} +1\)

Re-arrange: \(\sqrt{3-2x}-1 = \sqrt{2x}\);

Square: \((3-2x)-2\sqrt{3-2x}+1=2x\);

Re-arrange and reduce by 2: \(\sqrt{3-2x}=2-2x\);

Square: \(3-2x=4-8x+4x^2\)

\(4x^2=6x-1\).

Hope it's clear.

Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I hope you understand what I'm asking.

Thanks for any help!

First of all please read this:

rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

Next, if you square \(1-\sqrt{2x}=2x\) you'll get \(1 - 2\sqrt{2x}+2x=4x^2\). As you can see we don't have this answer among the options, so we should have done the other way around.

There are two ways of solving this given above:
if-root-3-2x-root-2x-1-then-4x-135539.html#p1104029
if-root-3-2x-root-2x-1-then-4x-135539.html#p1218265

sorry but a noob question here, i thought we aren't allowed to square both sides like this?

but instead all we can do is just to multiply both sides or divide both sides by the same number/objects etc.

cause when you square both sides, both sides is multiplying with different things right???

this part i don't get.

Hi melaos,

You are right when you say that we need to multiply or divide by the same number/object on both sides of an equation. Let me tell you why squaring both sides of an equation is a perfectly acceptable way.

Consider an equation a = b. This equation tells us that the value of a is the same as value of b. When we square a number, we multiply the number by itself. So, squaring this equation would look like a*a = b* b. Now, you would be tempted to say that we are multiplying different numbers on both sides of the equation.

Before you say so, look back at our original equation which says a = b. Hence, when we multiply a on LHS and b on RHS we are multiplying both sides by variables which have the same values.
Hence, squaring both sides of an equation is allowed.

Hope its clear!

Regards
Harsh

Asked: If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

\(3 - 2x = 2x + 1 + 2\sqrt{2x}\)
\(2 - 4x = 2\sqrt{2x}\)
\((1-2x)^2 = 4x^2 + 1 - 4x = 2x\)
\(4x^2 = 6x - 1\)

IMO E

Bunuel.. Why u r not proceeding here with modulus while solving square root.
As square of (3-2x)^(1/2) should be written as mod(3-2x)..??

Please heldp

\(\sqrt{a^2} = |a|\). Do we have the square of an expression in \(\sqrt{3-2x}\)? No. Hence, \(\sqrt{3-2x}\) does not equal to \(|3-2x|\). If it were \(\sqrt{(3-2x)^2}\), then it would be equal to \(|3-2x|\).

Bunuel.. Got it.. Thankyou so much for clarification..
You are always a saviour..

3-2x=2x+1+2rt(2x)
2-4x=2rt(2x)
4+16x^2-16x=8x
16x^2=24x-4
4x^2=6x-1