To show algebraically that there is no value of [tex]\( x \)[/tex] for which [tex]\( p(x) = q(x) \)[/tex], we need to set the given functions equal to each other and solve for [tex]\( x \)[/tex]. The functions are given by:
[tex]\[ p(x) = -2(x + 1) \][/tex]
[tex]\[ q(x) = x^2 - 5x + 2 \][/tex]
Step 1: Set [tex]\( p(x) \)[/tex] equal to [tex]\( q(x) \)[/tex]:
[tex]\[ -2(x + 1) = x^2 - 5x + 2 \][/tex]
Step 2: Simplify the left-hand side:
[tex]\[ -2x - 2 = x^2 - 5x + 2 \][/tex]
Step 3: Move all terms to one side of the equation to set it to zero:
[tex]\[ 0 = x^2 - 5x + 2 + 2x + 2 \][/tex]
Step 4: Combine like terms:
[tex]\[ 0 = x^2 - 3x + 4 \][/tex]
Step 5: Now, we need to solve the quadratic equation [tex]\( x^2 - 3x + 4 = 0 \)[/tex]. This quadratic equation can be solved using the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 4 \)[/tex].
Step 6: Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
Step 7: Simplify the terms inside the square root:
[tex]\[ x = \frac{3 \pm \sqrt{9 - 16}}{2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{-7}}{2} \][/tex]
Step 8: Notice that the term inside the square root is negative ([tex]\( -7 \)[/tex]), which means that the solutions for [tex]\( x \)[/tex] are complex numbers. The square root of a negative number introduces the imaginary unit [tex]\( i \)[/tex], where [tex]\( i = \sqrt{-1} \)[/tex].
[tex]\[ x = \frac{3 \pm \sqrt{7}i}{2} \][/tex]
Hence, the solutions are:
[tex]\[ x = \frac{3}{2} - \frac{\sqrt{7}}{2}i \][/tex]
[tex]\[ x = \frac{3}{2} + \frac{\sqrt{7}}{2}i \][/tex]
These solutions are complex numbers, meaning they involve imaginary components. Since we are only interested in real numbers and these solutions include imaginary parts, there are no real values of [tex]\( x \)[/tex] for which [tex]\( p(x) = q(x) \)[/tex]. This confirms that there is no value of [tex]\( x \)[/tex] in the real numbers for which [tex]\( p(x) = q(x) \)[/tex].
