Question

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

Answer 1

Answer:

Given the equations:

$y = \sqrt{1-x^2}$           ....[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 $x = \frac{4}{5} = 0.8$

For x = 0,

$y=\sqrt{1-0} = 1$

y = 2(0)-1 = -1

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

Therefore, the values of x is 0.8