Question - Solving Quadratic Equations for Specific Function Values
mathsolving quadratic equationsquadratic formularoots of quadratic equationsspecific function values
Solution:
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 $$ h(x) = -5 $$.
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