Question

5. Find the roots for each quadratic function
a) f(x) = 3x2 - 16x + 20
b) f(x) = 2x2-x -10
c) f(x) = 4x2 + 2x -3
6. Convert each quadratic equation to standard form
a) f(x) = -2(x + 5)(x - 7)
b) f(x) = (2x - 5)(x + 9)
c).
f(x) = 3(x+3)2+7
d) f(x) = -4(x - 1)≥ - 8

Answer 1

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:
1. \( x = \frac{20}{6} = \frac{10}{3} \)
2. \( x = \frac{12}{6} = 2 \)

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:
1. \( x = \frac{10}{4} = \frac{5}{2} \)
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).