Answer :
To solve the inequality [tex]\( 8x^2 - 8x + 12 < 9 \)[/tex], let's follow these steps:
1. Rearrange the inequality:
We first want to move all terms to one side of the inequality to set it to zero on the other side.
[tex]\[ 8x^2 - 8x + 12 - 9 < 0 \][/tex]
This simplifies to:
[tex]\[ 8x^2 - 8x + 3 < 0 \][/tex]
2. Determine the quadratic expression:
The expression we need to consider is:
[tex]\[ 8x^2 - 8x + 3 \][/tex]
3. Find the roots of the quadratic equation:
To determine where the quadratic equals zero, we solve:
[tex]\[ 8x^2 - 8x + 3 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 8 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 8 \cdot 3}}{2 \cdot 8} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{8 \pm \sqrt{64 - 96}}{16} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{-32}}{16} \][/tex]
Since the term under the square root, [tex]\(\sqrt{-32}\)[/tex], is a negative number, the quadratic equation has no real roots. In other words, the expression [tex]\(8x^2 - 8x + 3\)[/tex] cannot be factored into real numbers because it does not cross the x-axis.
4. Analyze the quadratic expression:
Since the quadratic [tex]\( 8x^2 - 8x + 3 \)[/tex] does not cross the x-axis, it is always positive or always negative. To determine which, consider the coefficient of [tex]\( x^2 \)[/tex]:
- The coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 8 \)[/tex], which is positive.
This means the quadratic always opens upwards, indicating that the expression [tex]\( 8x^2 - 8x + 3 \)[/tex] is always positive for all real values of [tex]\( x \)[/tex].
5. Conclude the inequality:
Given that the quadratic expression [tex]\( 8x^2 - 8x + 3 \)[/tex] is always positive and we are looking for where it is less than 0, there are no real solutions to the inequality:
[tex]\[ 8x^2 - 8x + 3 < 0 \][/tex]
Therefore, the inequality [tex]\( 8x^2 - 8x + 12 < 9 \)[/tex] has no real solution. The inequality never holds true for any real number [tex]\( x \)[/tex].
1. Rearrange the inequality:
We first want to move all terms to one side of the inequality to set it to zero on the other side.
[tex]\[ 8x^2 - 8x + 12 - 9 < 0 \][/tex]
This simplifies to:
[tex]\[ 8x^2 - 8x + 3 < 0 \][/tex]
2. Determine the quadratic expression:
The expression we need to consider is:
[tex]\[ 8x^2 - 8x + 3 \][/tex]
3. Find the roots of the quadratic equation:
To determine where the quadratic equals zero, we solve:
[tex]\[ 8x^2 - 8x + 3 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 8 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 8 \cdot 3}}{2 \cdot 8} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{8 \pm \sqrt{64 - 96}}{16} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{-32}}{16} \][/tex]
Since the term under the square root, [tex]\(\sqrt{-32}\)[/tex], is a negative number, the quadratic equation has no real roots. In other words, the expression [tex]\(8x^2 - 8x + 3\)[/tex] cannot be factored into real numbers because it does not cross the x-axis.
4. Analyze the quadratic expression:
Since the quadratic [tex]\( 8x^2 - 8x + 3 \)[/tex] does not cross the x-axis, it is always positive or always negative. To determine which, consider the coefficient of [tex]\( x^2 \)[/tex]:
- The coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 8 \)[/tex], which is positive.
This means the quadratic always opens upwards, indicating that the expression [tex]\( 8x^2 - 8x + 3 \)[/tex] is always positive for all real values of [tex]\( x \)[/tex].
5. Conclude the inequality:
Given that the quadratic expression [tex]\( 8x^2 - 8x + 3 \)[/tex] is always positive and we are looking for where it is less than 0, there are no real solutions to the inequality:
[tex]\[ 8x^2 - 8x + 3 < 0 \][/tex]
Therefore, the inequality [tex]\( 8x^2 - 8x + 12 < 9 \)[/tex] has no real solution. The inequality never holds true for any real number [tex]\( x \)[/tex].