Let's solve the given problems one by one.
Problem 5: Find the roots for each quadratic function
a) \( f(x) = 3x^2 - 16x + 20 \)
To find the roots, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = -16 \), and \( c = 20 \).
First, calculate the discriminant: \[ b^2 - 4ac = (-16)^2 - 4(3)(20) = 256 - 240 = 16 \]
Now, apply the quadratic formula: \[ x = \frac{16 \pm \sqrt{16}}{6} = \frac{16 \pm 4}{6} \]
This gives two solutions:
- \( x = \frac{20}{6} = \frac{10}{3} \)
- \( x = \frac{12}{6} = 2 \)
So the roots are \( x = \frac{10}{3} \) and \( x = 2 \).
b) \( f(x) = 2x^2 - x - 10 \)
Again, we apply the quadratic formula where \( a = 2 \), \( b = -1 \), and \( c = -10 \):
Calculate the discriminant: \[ b^2 - 4ac = (-1)^2 - 4(2)(-10) = 1 + 80 = 81 \]
Now using the formula: \[ x = \frac{-(-1) \pm \sqrt{81}}{2(2)} = \frac{1 \pm 9}{4} \]
This gives two solutions:
- \( x = \frac{10}{4} = \frac{5}{2} \)
- \( x = \frac{-8}{4} = -2 \)
So the roots are \( x = \frac{5}{2} \) and \( x = -2 \).
c) \( f(x) = 4x^2 + 2x - 3 \)
Using the quadratic formula where \( a = 4 \), \( b = 2 \), and \( c = -3 \):
Calculate the discriminant: \[ b^2 - 4ac = (2)^2 - 4(4)(-3) = 4 + 48 = 52 \]
Now find the roots: \[ x = \frac{-2 \pm \sqrt{52}}{2(4)} = \frac{-2 \pm 2\sqrt{13}}{8} = \frac{-1 \pm \sqrt{13}}{4} \]
So the roots are \( x = \frac{-1 + \sqrt{13}}{4} \) and \( x = \frac{-1 - \sqrt{13}}{4} \).
Problem 6: Convert each quadratic equation to standard form
a) \( f(x) = -2(x + 5)(x - 7) \)
Expanding this: \[ f(x) = -2[(x)(x) + (7)(x) + (5)(x) - (35)] = -2[x^2 - 2x - 35] = -2x^2 + 4x + 70 \]
So, the standard form is \( f(x) = -2x^2 + 4x + 70 \).
b) \( f(x) = (2x - 5)(x + 9) \)
Expanding this: \[ f(x) = (2x)(x) + (2x)(9) - (5)(x) - (5)(9) = 2x^2 + 18x - 5x - 45 = 2x^2 + 13x - 45 \]
So, the standard form is \( f(x) = 2x^2 + 13x - 45 \).
c) \( f(x) = 3(x + 3)^2 + 7 \)
Expanding this: \[ f(x) = 3[(x^2 + 6x + 9)] + 7 = 3x^2 + 18x + 27 + 7 = 3x^2 + 18x + 34 \]
So, the standard form is \( f(x) = 3x^2 + 18x + 34 \).
d) \( f(x) = -4(x - 1) \geq -8 \)
First, rewrite \( f(x) = -4(x - 1) \): \[ f(x) = -4x + 4 \quad \text{and set it equal to } -8. \]
To rewrite this function as a standard form: \[ f(x) = -4x + 4 \]
So, for the standard form of \( f(x) \), we have \( f(x) = -4x + 4 \).
In summary:
Roots of Quadratics: a) \( x = \frac{10}{3}, 2 \)
b) \( x = \frac{5}{2}, -2 \)
c) \( x = \frac{-1 + \sqrt{13}}{4}, \frac{-1 - \sqrt{13}}{4} \)
Standard Forms: a) \( -2x^2 + 4x + 70 \)
b) \( 2x^2 + 13x - 45 \)
c) \( 3x^2 + 18x + 34 \)
d) \( -4x + 4 \) (as an equation, but it can be considered standard form as linear).