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Chapter 1: Problem 48

Find the value(s) of \(x\) for which \(f(x)=g(x)\). $$f(x)=\sqrt{x}-4, \quad g(x)=2-x$$

Short Answer

Expert verified
There are no real values of \(x\) for which \(f(x) = g(x)\). The equation has no real solutions since the term under the square root in the quadratic formula is negative.

Step by step solution

01

Set the functions equal to each other

To find the value of \(x\) at which \(f(x) = g(x)\), set \(\sqrt{x}-4\) equal to \(2-x\), i.e., \(\sqrt{x}-4 = 2-x\).
02

Rearrange the equation

To avoid having a negative term on the right side and to deal with the square root on the left side, rearrange the equation to \(\sqrt{x} = 4 - 2 + x\) which simplifies to \(\sqrt{x} = x+2\).
03

Square both sides of the equation

To remove the square root, square both sides of the equation, resulting in \(\sqrt{x}^2 = (x+2)^2\), or \(x = x^2 + 4x + 4\).
04

Rearrange the equation to set it to zero

To solve for \(x\), rearrange the equation to \(x^2 + 3x + 4 = 0\)
05

Solve for \(x\) using the quadratic formula

The quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), can be used to find the solutions. Here, \(a = 1\), \(b = 3\), and \(c = 4\). However, the term under the square root, \(b^{2}-4ac = 9 - 16 = -7\) is negative. This means there are no real solutions for \(x\), given the two functions.

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