Answer :
Certainly! Let's solve the problem in two parts.
### Part a: Express [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex]
We are asked to express the quadratic expression [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex].
1. Factor out the coefficient of [tex]\( x^2 \)[/tex] from the quadratic portion:
[tex]\[ 9x^2 - 15x = 9(x^2 - \frac{15}{9} x) \][/tex]
Simplify the fraction:
[tex]\[ 9(x^2 - \frac{5}{3} x) \][/tex]
2. Complete the square inside the parentheses:
- Take the coefficient of [tex]\( x \)[/tex], [tex]\(-\frac{5}{3}\)[/tex], and halve it:
[tex]\[ \frac{-\frac{5}{3}}{2} = -\frac{5}{6} \][/tex]
- Square the result:
[tex]\[ \left( -\frac{5}{6} \right)^2 = \frac{25}{36} \][/tex]
3. Add and subtract this value inside the parentheses, and then factor:
[tex]\[ 9 \left( x^2 - \frac{5}{3} x + \frac{25}{36} - \frac{25}{36} \right) \][/tex]
[tex]\[ = 9 \left( (x - \frac{5}{6})^2 - \frac{25}{36} \right) \][/tex]
4. Distribute the 9:
[tex]\[ 9 \left( (x - \frac{5}{6})^2 \right) - 9 \left( \frac{25}{36} \right) \][/tex]
Simplify the second term:
[tex]\[ 9 \left( (x - \frac{5}{6})^2 \right) - \frac{225}{36} = 9 \left( (x - \frac{5}{6})^2 \right) - \frac{25}{4} \][/tex]
Therefore, the expression [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex] is:
[tex]\[ (3x - \frac{5}{2})^2 - \frac{25}{4} \][/tex]
Here, [tex]\( a = \frac{5}{2} \)[/tex] and [tex]\( b = \frac{25}{4} \)[/tex].
### Part b: Solve the inequality [tex]\( 9x^2 - 15x < 6 \)[/tex]
We need to find the values of [tex]\( x \)[/tex] that satisfy the inequality:
[tex]\[ 9x^2 - 15x < 6 \][/tex]
First, rearrange the inequality to have [tex]\( 0 \)[/tex] on one side:
[tex]\[ 9x^2 - 15x - 6 < 0 \][/tex]
Now, we need to find the roots of the equation [tex]\( 9x^2 - 15x - 6 = 0 \)[/tex] to determine the intervals to test. Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 9 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(9)(-6)}}{2(9)} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{225 + 216}}{18} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{441}}{18} \][/tex]
[tex]\[ x = \frac{15 \pm 21}{18} \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{36}{18} = 2 \quad \text{and} \quad x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
The quadratic [tex]\( 9x^2 - 15x - 6 \)[/tex] opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive), so the inequality [tex]\( 9x^2 - 15x - 6 < 0 \)[/tex] is satisfied between the roots [tex]\( -\frac{1}{3} \)[/tex] and [tex]\( 2 \)[/tex].
Therefore, the set of values of [tex]\( x \)[/tex] that satisfy the inequality is:
[tex]\[ -\frac{1}{3} < x < 2 \][/tex]
So, the solution is:
[tex]\[ x \in \left( -\frac{1}{3}, 2 \right) \][/tex]
### Part a: Express [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex]
We are asked to express the quadratic expression [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex].
1. Factor out the coefficient of [tex]\( x^2 \)[/tex] from the quadratic portion:
[tex]\[ 9x^2 - 15x = 9(x^2 - \frac{15}{9} x) \][/tex]
Simplify the fraction:
[tex]\[ 9(x^2 - \frac{5}{3} x) \][/tex]
2. Complete the square inside the parentheses:
- Take the coefficient of [tex]\( x \)[/tex], [tex]\(-\frac{5}{3}\)[/tex], and halve it:
[tex]\[ \frac{-\frac{5}{3}}{2} = -\frac{5}{6} \][/tex]
- Square the result:
[tex]\[ \left( -\frac{5}{6} \right)^2 = \frac{25}{36} \][/tex]
3. Add and subtract this value inside the parentheses, and then factor:
[tex]\[ 9 \left( x^2 - \frac{5}{3} x + \frac{25}{36} - \frac{25}{36} \right) \][/tex]
[tex]\[ = 9 \left( (x - \frac{5}{6})^2 - \frac{25}{36} \right) \][/tex]
4. Distribute the 9:
[tex]\[ 9 \left( (x - \frac{5}{6})^2 \right) - 9 \left( \frac{25}{36} \right) \][/tex]
Simplify the second term:
[tex]\[ 9 \left( (x - \frac{5}{6})^2 \right) - \frac{225}{36} = 9 \left( (x - \frac{5}{6})^2 \right) - \frac{25}{4} \][/tex]
Therefore, the expression [tex]\( 9x^2 - 15x \)[/tex] in the form [tex]\( (3x - a)^2 - b \)[/tex] is:
[tex]\[ (3x - \frac{5}{2})^2 - \frac{25}{4} \][/tex]
Here, [tex]\( a = \frac{5}{2} \)[/tex] and [tex]\( b = \frac{25}{4} \)[/tex].
### Part b: Solve the inequality [tex]\( 9x^2 - 15x < 6 \)[/tex]
We need to find the values of [tex]\( x \)[/tex] that satisfy the inequality:
[tex]\[ 9x^2 - 15x < 6 \][/tex]
First, rearrange the inequality to have [tex]\( 0 \)[/tex] on one side:
[tex]\[ 9x^2 - 15x - 6 < 0 \][/tex]
Now, we need to find the roots of the equation [tex]\( 9x^2 - 15x - 6 = 0 \)[/tex] to determine the intervals to test. Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 9 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(9)(-6)}}{2(9)} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{225 + 216}}{18} \][/tex]
[tex]\[ x = \frac{15 \pm \sqrt{441}}{18} \][/tex]
[tex]\[ x = \frac{15 \pm 21}{18} \][/tex]
Thus, the roots are:
[tex]\[ x = \frac{36}{18} = 2 \quad \text{and} \quad x = \frac{-6}{18} = -\frac{1}{3} \][/tex]
The quadratic [tex]\( 9x^2 - 15x - 6 \)[/tex] opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive), so the inequality [tex]\( 9x^2 - 15x - 6 < 0 \)[/tex] is satisfied between the roots [tex]\( -\frac{1}{3} \)[/tex] and [tex]\( 2 \)[/tex].
Therefore, the set of values of [tex]\( x \)[/tex] that satisfy the inequality is:
[tex]\[ -\frac{1}{3} < x < 2 \][/tex]
So, the solution is:
[tex]\[ x \in \left( -\frac{1}{3}, 2 \right) \][/tex]
