The square root of 1 less than twice a number is equal to 2 less than the number.Find the number.
I read that as:
√(2x - 1) = x-2
square both sides
2x-1 = x^2 - 4x + 4
x^2 - 6x + 5 = 0
(x-5)(x-1) = 0
x = 5 or x = 1
since I squared, both answers must be verified in the original equation,
if x = 1
LS = √(2x-1) = √1 = 1
RS = 1-2 = -1
so x ≠ 1
if x = 5
LS = √(10-1) = 3
RS = 5-2 = 3
your number is 5
Let's solve this step-by-step.
Step 1: Let's assume the number is "x".
Step 2: It says "1 less than twice a number". So, we have 2x - 1.
Step 3: It says "the square root of 1 less than twice a number". So, we have the square root of (2x - 1).
Step 4: It says "is equal to 2 less than the number". So, we have (2x - 1) = x - 2.
Step 5: Let's solve the equation we got in Step 4.
Moving all the "x" terms to one side: (2x - x) = -1 - 2
Simplifying: x = -3
Step 6: Therefore, the number is -3.
So, the number is -3.
To find the number, let's break down the problem step by step:
Let the number be represented by x.
According to the problem, the square root of (1 less than twice the number) is equal to (2 less than the number).
Step 1: Translate the given problem into an equation:
The square root of (2x - 1) = (x - 2)
Step 2: Square both sides to eliminate the square root:
(√(2x - 1))^2 = (x - 2)^2
Simplifying the equation:
2x - 1 = (x - 2)(x - 2)
Step 3: Expand the equation:
2x - 1 = x^2 - 4x + 4
Step 4: Rearrange the equation to set it equal to zero:
x^2 - 6x + 5 = 0
Step 5: Factor the quadratic equation:
(x - 5)(x - 1) = 0
Step 6: Set each factor equal to zero and solve for x:
x - 5 = 0
x = 5
x - 1 = 0
x = 1
Therefore, the possible values for x are 5 and 1.