If √x²= 2x + 3, then x=?

Question

show ur working n step by step explanation

Answer 1

x=-1

Proof:
Squaring both sides of the equation, we get
x²=4x²+12x+9.
Then subtracting both sides by x²,
0=3x²+12x+9.
Since the resulting equation is factorable by 3, then
0=x²+4x+3.
Factoring the equation will give
0=(x+3)(x+1).
Then,
x+3=0 and x+1=0
x=-3 and x=-1
But we want 2x+3>0 so that it will satisfy √x² that is a positive number. Then, when x=-3,
√(-3)²=2(-3)+3
3 = -3 which is not true.
When x=-1,
√(-1)²=2(-1)+3
1 = 1 which is true.
Therefore, x=-1

Answer 2

FIRST SQUARE BOTH SIDES TO ELIMINATE THE √ SIGN
(√x^2)=(2x+3)^2
x^2=(2x+3)^2
maintain what is on the left side of the = sign and expand what is on the right hand side. when expanding, take one term in the () and multiply it by all the terms in the ().
x^2=2x(2x+3)+3(2x+3)
again maintain what is on the left side of the = sign and expand what is on the right hand side.
x^2=4x^2+6x+6x+9
then group llike terms
x^2=4x^2+12x+9
send what is on the LHS to the RHS and equate it to 0
4x^2+12x+9-x^2=0
note very well that when x^2 crosses the= sign it becomes -x^2
group like terms again
4x^2-x^2+12x+9=0
3x^2+12x+9=0
you can still simplfy it by divividing all the terms in the equation by 3
x^2+4x+3=0
note that 0/3 =0
u will find 2 numbers when u multiply them u will get the constant which is 3 and at the same time when u add them u will get the co-efficint of x which is 4. the numbers should be factors of the constant which is 3. those 2 numbers are 1and 3.
u then replace the 4x with x and 3. NB 1x=x
x^2+x+3x+3=0
(x^2+x)+(3x+3)=0
then factorize them.
x(x+1)+3(x+1)=0
u can see that all those in the () are equal so we take one of them and then take the others outside the () and multiply them.
(x+1)(x+3)=0
we take them one after the other and solve for x
x+1=0 x+3=0
x=-1 x=-3
x=-1 or -3

Answer 3

To remove the square root from the left side of the equal sign, we square the right hand side of the equal sign, we get

x^2 = (2x +3)^2
or that is x^2 = 4x^2 + 12x + 9 [applying the formula (a+b)^2 = a^2 + 2ab + b^2]
or that is, 4x^2 - x^2 +12x +9 = 0
or that is, 3x^2 +12x +9 =0

This is is quadratic equation and so we solve for quadratic equations. We get

or that is, 3x^2 + 3x +9x +9 = 0
3x (x +1 ) + 9 (x +1) = 0
(x + 1) (3x +9) =0
hence, either x+1 =0 viz. x = -1
or 3x +9 =0 viz x = -9/3 = -3

Therefore, x = -1 or x= -3

Answer 4

√x² = | x | and not x.
eg. √(-3)² = √9 = 3 and not -3.

:. | x | = 2x + 3

Case 1: x >= 0

x = 2x + 3
x = -3
:. for x >= 0 has no solution as x = -3 is less than 0

Case 2: x < 0

-x = 2x + 3
3x = -3
x = -1
:. x = -1 for set x < 0



:. x = -1 is the only answer.

Answer 5

Taking square on both sides, we get

x^2 = (2x +3)^2
x^2 = 4x^2 + 12x + 9
4x^2 - x^2 +12x +9 = 0
3x^2 +12x +9 =0
3x^2 + 3x +9x +9 = 0
3x (x +1 ) + 9 (x +1) = 0
(x + 1) (3x +9) =0
x+1 =0 or 3x +9 =0

x= -1 or x= -3

solution set for x is { -1, -3 }

Answer 6

We know that sqrt(x^2) is equivalent to +/ - abs(x), where
abs(x)
= x, whenever x>0
= - x, whenever, x< 0
= 0, whenever x=0
So, given equation becomes:

Case I: x=0, sqrt(x^2) = +/- abs(x)
0 = 2*0+3, for x=0
Or, 0 = 3, which is a contradiction, so that x cannot be 0.

Case 2: x> 0, sqrt(x^2) = + abs(x)
x = 2x+3,
Or, x = -3, which is a contradiction.

Case 3: x<0, sqrt(x^2) = + abs(x)

-x = 2x+3
Or, x =1, which is a contradiction.

Case 4: x> 0, sqrt(x^2) = - abs(x)
-x = 2x+3,
Or, x = 1.

Case 5: x<0, sqrt(x^2) = - abs(x)

x = 2x+3
Or, x = -3.
Hence, comparing all the five cases, x = 1, -3.

Answer 7

then,
x^2 = (2x + 3 )^2
x^2 = 4x^2 + 12x +9
0 = 3x^ + 12x + 9
0 = 3x^2 + 3x + 9x +9
0 = 3x(x+1)+9(x+1)
0 = (x+1)(3x+9)
Either,
x+1=0
x = -1
Or,
3x+9 = 0
x = -3.
This is the solution.
But i think , homeworks are given to be one by yourself.

Answer 8

2x+3

Answer 9

Answer is -1 or -3

sq rt of x^2 is +x or -x
Thus if +x = 2x+3
then x+3=0
or, x= -3
If -x=2x+3
then 3x+3=0
or 3x= -3
or x= -1

Answer 10

This Sol. Is The Most Precise Of All abd still
see its correct!!!!!
√x2=2x+3
=>x2=(2x+3)2
=>x=(2x+3) or =>x=-(2x+3)
=>x=-3 or =>x=-1
mind it!!!!!!!!