how do i solve for x here by using the square root method?(2x + 5)² = (x - 1)²?

Answer 1

(2x + 5)^2 = (x - 1)^2

Key note: Whenever you take the square root of both sides of an equation, you have to add a "plus or minus" on the right hand side. This is what the "square root" method is; the addition of the "+/-", and the three people above me failed to recognize that (until some recent editing). That is,

2x + 5 = +/- (x - 1)

This leads to two equations:

2x + 5 = x - 1
2x + 5 = -(x - 1)

2x + 5 = x - 1 means
x = -6

and 2x + 5 = -(x - 1) means
2x + 5 = -x + 1, so
3x = -4, x = -4/3

So x = {-6, -4/3}

Let's verify:

If x = -4/3, then

LHS = (2(-4/3) + 5)^2
LHS = (-8/3 + 5)^2
LHS = (-8/3 + 15/3)^2
LHS = (7/3)^2 = 49/9

RHS = (-4/3 - 1)^2
RHS = (-4/3 - 3/3)^2
RHS = (-7/3)^2 = 49/9

LHS = RHS, so x = -7/3 is a definite answer.

Test x = -6. Then
LHS = (2(-6) + 5)^2 = (-12 + 5)^2 = (-7)^2 = 49
RHS = (-6 - 1)^2 = (-7)^2 = 49

LHS = RHS, so x = -6 is another definite answer.

Answer 2

First: eliminate the exponents > find the square root of both sides

V`(2x + 5)^2 = V`(x - 1)^2

V`(2x+5)(2x+5) = V`(x - 1)(x - 1)

*Since you have two of the same expressions (the expression are repeated twice), pull out one of the expressions on both sides....

2x + 5 = x - 1

Sec: subtract 5 from both sides...

2x + 5 - 5 = x - 1 - 5
2x = x - 6

*Subtract "x" from both sides...

2x - x = x - x - 6
x = - 6

Answer 3

(2x+5)^2 = x^2-2x+1
2x+5 = square root of x^2-2x+1
2x+5 = x - square root of 2x +1
2x-x+square root of 2x = 1-5
x+square root of2x = -4
x+1.414x = -4
2.1414x = -4
x = -4/2.1414
= -1.87

Answer 4

(2x + 5)² = (x - 1)²

2X + 5 = square root of (x – 1)2

2X + 5 = x – 1

2x – x = -1 – 5

X = -6

Answer 5

Its just by square-rooting them
2x+5=x-1
=>2x-x=-1-5
=>x=-6
k

Answer 6

hey this is the perfect soln:

(2x + 5)² = (x - 1)² a²-b²=(a+b)(a-b)
= (2x + 5)² - (x - 1)²=0
= (2x+5+x-1)(2x+5-x+1)=0
= (3x+4)(x+6)=0
hence ur X is -4/3 or x=-6
njoy..

Answer 7

Soooooooo Easyyyyyyyyyy !!!
(2x + 5)^2 = (x -1)^2
4x^2 + 20x + 25 = x^2 - 2x + 1
3x^2 + 22x +24 = 0
x1 = -4/3
x2 = -6

Answer 8

follow this:
(2x+5)^2-(x-1)^2=0 => ((2x+5)+(x-1))*((2x+5)-(x-1))=
(3x-4)*(x+6)=0
=>x=4/3 or -6

Answer 9

taking sqrt

2x+5=-+-(x-1) So you have two cases

2x+5=x-1===> x=-6

2x+5= -x+1 ===> 3x=-4 and x=-4/3

both values solve the problem

Answer 10

sqrt of [((2x + 5)^2)=((x - 1)^2)]
2x + 5 = x - 1
2x - x = -1 - 5
x = -6