Answer :
To determine the exact value(s) for which the rational expression [tex]\( f(x) = \frac{-11x + 1}{-30x^2 - 25x + 5} \)[/tex] is undefined, we need to identify the values of [tex]\( x \)[/tex] that make the denominator zero because division by zero is undefined in mathematics.
Let's set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -30x^2 - 25x + 5 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = -30 \)[/tex], [tex]\( b = -25 \)[/tex], and [tex]\( c = 5 \)[/tex]. To find the roots of this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-25)^2 - 4(-30)(5) \][/tex]
[tex]\[ \Delta = 625 + 600 \][/tex]
[tex]\[ \Delta = 1225 \][/tex]
Since the discriminant is positive, we will have two distinct real roots. We proceed by substituting the values into the quadratic formula:
[tex]\[ x = \frac{{-(-25) \pm \sqrt{1225}}}{2(-30)} \][/tex]
[tex]\[ x = \frac{{25 \pm 35}}{-60} \][/tex]
Now, let's find the two roots:
1. For the [tex]\( + \)[/tex] case:
[tex]\[ x = \frac{{25 + 35}}{-60} = \frac{60}{-60} = -1 \][/tex]
2. For the [tex]\( - \)[/tex] case:
[tex]\[ x = \frac{{25 - 35}}{-60} = \frac{-10}{-60} = \frac{1}{6} \][/tex]
Thus, the values of [tex]\( x \)[/tex] that make the denominator zero and the rational expression [tex]\( f(x) \)[/tex] undefined are:
[tex]\[ x = -1, \frac{1}{6} \][/tex]
Therefore, the rational expression [tex]\( f(x) = \frac{-11x + 1}{-30x^2 - 25x + 5} \)[/tex] is undefined for [tex]\( x = -1 \)[/tex] and [tex]\( x = \frac{1}{6} \)[/tex].
Providing the answer in the required format,
[tex]\[ x = -1, \frac{1}{6} \][/tex]
Let's set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -30x^2 - 25x + 5 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = -30 \)[/tex], [tex]\( b = -25 \)[/tex], and [tex]\( c = 5 \)[/tex]. To find the roots of this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-25)^2 - 4(-30)(5) \][/tex]
[tex]\[ \Delta = 625 + 600 \][/tex]
[tex]\[ \Delta = 1225 \][/tex]
Since the discriminant is positive, we will have two distinct real roots. We proceed by substituting the values into the quadratic formula:
[tex]\[ x = \frac{{-(-25) \pm \sqrt{1225}}}{2(-30)} \][/tex]
[tex]\[ x = \frac{{25 \pm 35}}{-60} \][/tex]
Now, let's find the two roots:
1. For the [tex]\( + \)[/tex] case:
[tex]\[ x = \frac{{25 + 35}}{-60} = \frac{60}{-60} = -1 \][/tex]
2. For the [tex]\( - \)[/tex] case:
[tex]\[ x = \frac{{25 - 35}}{-60} = \frac{-10}{-60} = \frac{1}{6} \][/tex]
Thus, the values of [tex]\( x \)[/tex] that make the denominator zero and the rational expression [tex]\( f(x) \)[/tex] undefined are:
[tex]\[ x = -1, \frac{1}{6} \][/tex]
Therefore, the rational expression [tex]\( f(x) = \frac{-11x + 1}{-30x^2 - 25x + 5} \)[/tex] is undefined for [tex]\( x = -1 \)[/tex] and [tex]\( x = \frac{1}{6} \)[/tex].
Providing the answer in the required format,
[tex]\[ x = -1, \frac{1}{6} \][/tex]
