Question

At which approximate x-value are these two equations equal
y = √1 - x^2
y = 2x - 1

Answer 1

Answer: on Plato it should be .8 or b

Explanation:

Answer 2

Given the equations:

`y = √(1-x^2)` ....[1]

`y = 2x-1` .....[2]

Taking square both sides in [1] we have;

`y^2 = 1-x^2` ....[3]

Substitute the equation [2] into [3] we have;

`(2x-1)^2 = 1-x^2`

Using identity rule:
`(a-b)^2 = a^-2ab+b^2`

then;

`4x^2-4x+1 = 1-x^2`

Add
`x^2` both sides we have;

`5x^2-4x+1 = 1`

Subtract 1 from both sides we have;

`5x^2-4x=0`

⇒
`x(5x-4) = 0`

By zero product property we have;

x = 0 and 5x-4 = 0

⇒x= 0 and
`x = (4)/(5) = 0.8`

For x = 0,

`y=√(1-0) = 1`

y = 2(0)-1 = -1

Since, the value of x = 0 doesn't satisfy the equations.

Therefore, the values of x is 0.8