Solving Quadratic Equations for Specific Function Values
The image shows a problem from a textbook asking to graph the function \( h(x) = x^2 - 6x + 5 \) and solve the equations for the following conditions:
a) \( h(x) = 5 \)
b) \( h(x) = 0 \)
c) \( h(x) = -5 \)
Let's solve these equations one by one.
a) To solve \( h(x) = 5 \), we substitute \( 5 \) for \( h(x) \) in the original function:
\( x^2 - 6x + 5 = 5 \)
Subtracting \( 5 \) from both sides of the equation gives:
\( x^2 - 6x = 0 \)
Factoring out an \( x \) gives:
\( x(x - 6) = 0 \)
So \( x = 0 \) or \( x = 6 \). These are the values of \( x \) when \( h(x)=5 \).
b) To solve \( h(x) = 0 \), we need to find the roots of the quadratic equation:
\( x^2 - 6x + 5 = 0 \)
Factoring the quadratic gives:
\( (x - 5)(x - 1) = 0 \)
Hence, the values of \( x \) that make \( h(x) = 0 \) are \( x = 5 \) and \( x = 1 \).
c) To solve \( h(x) = -5 \), we substitute \( -5 \) for \( h(x) \) in the original function:
\( x^2 - 6x + 5 = -5 \)
Adding \( 5 \) to both sides gives:
\( x^2 - 6x + 10 = 0 \)
This quadratic equation does not factor nicely, so we will need to use the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
where \( a = 1 \), \( b = -6 \), and \( c = 10 \). Plugging these values in gives:
\( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)} \)
\( x = \frac{6 \pm \sqrt{36 - 40}}{2} \)
\( x = \frac{6 \pm \sqrt{-4}}{2} \)
Since there is a negative number inside the square root, the solutions for \( x \) will be complex numbers. Hence, there are no real solutions for \( x \) when \( h(x) = -5 \).
