If a and b are the roots, (x - a)(x - b) = 0

For the roots $\frac{1}{2}$ and -3,

(x - $\frac{1}{2}$)(x - (-3)) = 0

-- = -(-) = +

(x - $\frac{1}{2}$)(x + 3) = 0

Expand the brackets

x(x + 3) - $\frac{1}{2}$(x + 3) = 0

x² + 3x - $\frac{\mathrm{x}}{2}$ - $\frac{3}{2}$ = 0

Get rid of the fractions by multiplying both sides by 2

2x² + 6x - x - 3 = 0

2x² + 5x - 3 = 0

Now compare the equation to the general equation to find each of the variables.

px² + qx + r = 0

p = 2
q = 5
r = -3

If α and β are the roots, then

(x - α)(x - β) = 0

(x - (-2q))(x - 5q) = 0

Note: -- = + or -(-) = +

(x + 2q)(x - 5q) = 0

x(x - 5q) + 2q(x - 5q) = 0

x² - 5qx + 2qx - 10q² = 0

x² - 3qx - 10q² = 0

If α and β are the roots, then

(x - α)(x - β) = 0

(x - $\frac{1}{2}$)(x - -$\frac{1}{3}$) = 0

Note: -- = +

(x - $\frac{1}{2}$)(x + $\frac{1}{3}$) = 0

x(x + $\frac{1}{3}$) - $\frac{1}{2}$(x + $\frac{1}{3}$) = 0

x² + $\frac{\mathrm{x}}{3}$ - $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 1}}{\mathrm{2\; x\; 3}}$ = 0

x² + $\frac{\mathrm{x}}{3}$ - $\frac{\mathrm{x}}{2}$ - $\frac{1}{6}$ = 0

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3, 2 and 6.

The L.C.M of 3, 2 and 6 is 6.

6 x x² + 6 x $\frac{\mathrm{x}}{3}$ - 6 x $\frac{\mathrm{x}}{2}$ - 6 x $\frac{1}{6}$ = 0 x 6

6x² + 2x - 3x - 1 = 0

6x² - x - 1 = 0

If (x + a) is a factor, then when you substitute -a into the equation, you should get 0.

Since (x + 2) is a factor of x² + px - 10, substituting -2 into x² + px - 10 should give 0.

(-2)² + p(-2) - 10 = 0

4 - 2p - 10 = 0

-6 = 2p

Divide both sides by 2

p = $\frac{-6}{2}$ = -3

4x² - 16x + 15 = 0

Notes:

1. a x c = 4 x 15 = 60

2. Split b (-16) such that two numbers when you sum you will get b (-16) and when you multiply you will get ac (60). The two numbers are -10 and -6

3. -16x = -10x - 6x

4x² -10x - 6x + 15 = 0

2x(2x - 5) - 3(2x - 5) = 0

(2x - 5)(2x - 3) = 0

When 2x - 5 = 0, x = $\frac{5}{2}$ = 2$\frac{1}{2}$

When 2x - 3 = 0, x = $\frac{3}{2}$ = 1$\frac{1}{2}$

x² - 5x - 14

Notes:

Split -5 in such a way that two numbers when you multiply you will get -14 and when you add you will get -5.

The numbers are -7 and 2

-5x = -7x + 2x

x² -7x + 2x - 14

x(x - 7) + 2(x - 7)

(x - 7)(x + 2)

x² - 9x + 14

Notes:

Split -9 in such a way that two numbers when you multiply you will get 14 and when you add you will get -9.

The numbers are -7 and -2

-9x = -7x - 2x

x² -7x - 2x + 14

x(x - 7) - 2(x - 7)

(x - 7)(x - 2)

$\frac{{\mathrm{x}}^2-\mathrm{5x}-14}{{\mathrm{x}}^2-\mathrm{9x}+14}$ = $\frac{\mathrm{(x\; -\; 7)(x\; +\; 2)}}{\mathrm{(x\; -\; 7(x\; -\; 2)}}$

(x - 7) cancels out.

$\frac{{\mathrm{x}}^2-\mathrm{5x}-14}{{\mathrm{x}}^2-\mathrm{9x}+14}$ = $\frac{\mathrm{x\; +\; 2}}{\mathrm{x\; -\; 2}}$

(4x² + 3x - 1)(3x + 1) = (4x² + 3x - 1)(3x + 1)

(4x² + 3x - 1)(3x + 1) = 3x(4x² + 3x - 1) + 1(4x² + 3x - 1)

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² - 3x + 4x² + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 9x² + 4x² - 3x + 3x - 1

(4x² + 3x - 1)(3x + 1) = 12x³ + 13x² + 0x - 1

Method I

If α and β are the roots, then

(x - α)(x - β) = 0

(x - $\frac{1}{2}$)(x - -$\frac{3}{2}$) = 0

Note: -- = +

(x - $\frac{1}{2}$)(x + $\frac{3}{2}$) = 0

x(x + $\frac{3}{2}$) -$\frac{1}{2}$(x + $\frac{3}{2}$) = 0

x² + $\frac{\mathrm{3x}}{2}$ - $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = 0

x² + $\frac{\mathrm{3x\; -\; x}}{2}$ - $\frac{3}{4}$ = 0

x² + $\frac{\mathrm{2x}}{2}$ - $\frac{3}{4}$ = 0

x² + x - $\frac{3}{4}$ = 0

Get rid of the fraction by multiplying both sides by the denominator 4.

4 x x² + 4 x x - $\frac{3}{4}$ x 4 = 0 x 4

4x² + 4x - 3 = 0

Method II

x² - (sum of roots)x + product of roots = 0

Sum of roots = $\frac{1}{2}$ + -$\frac{3}{2}$

Sum of roots = $\frac{1}{2}$ - $\frac{3}{2}$

Sum of roots = $\frac{\mathrm{1\; -\; 3}}{2}$ = $\frac{-2}{2}$ = -1

Product of roots = $\frac{1}{2}$ x - $\frac{3}{2}$ = - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = - $\frac{3}{4}$

x² - (-1)x + - $\frac{3}{4}$ = 0

-(-) or -- = +

x² + x - $\frac{3}{4}$ = 0

Get rid of the fraction by multiplying both sides by the denominator 4

4 x x² + 4 x x - $\frac{3}{4}$ x 4 = 0 x 4

4x² + 4x - 3 = 0

Use points on the graph to substitute the general equation y = ax² + bx + c.

Note: it is easier to use points that either x = 0 or y = 0

When x = -1, y = 0

0 = a(-1)² + b(-1) + c

a - b + c = 0 ------- eqn 1

When x = 2, y = 0

0 = a(2)² + b(2) + c

4a + 2b + c = 0 ------- eqn 2

When y = 2, x = 0

2 = a(0)² + b(0) + c

c = 2

Substitute c = 2 into ----- eqn 1

a - b + 2 = 0

a - b = -2 -------- eqn 3

Substitute c = 2 into ----- eqn 2

4a + 2b + 2 = 0

4a + 2b = - 2

Divide both sides by 2

2a + b = - 1 ------ eqn 4

Get rid of b. ----- eqn 3 + ----- eqn 4

3a = -3

Divide both sides by 3.

a = $\frac{-3}{3}$ = -1

Substitute a = -1 into ---- eqn 4

2(-1) + b = -1

-2 + b = -1

b = -1 + 2

b = 1

Substitute a = -1, b = 1 and c = 2 into the general equation y = ax² + bx + c

y = ax² + bx + c

y = (-1)x² + 1(x) + 2

y = -x² + x + 2

y = 2 + x - x²

If a and b are the roots, (x - a)(x - b) = 0

Change the mixed fraction to improper fraction.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

For the roots -$\frac{1}{2}$ and $\frac{3}{2}$,

(x --$\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

-- = +

(x + $\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

x(x - $\frac{3}{2}$) + $\frac{1}{2}$(x - $\frac{3}{2}$) = 0

x² - $\frac{\mathrm{3x}}{2}$ + $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = 0

x² + $\frac{\mathrm{x\; -\; 3x}}{2}$ - $\frac{3}{4}$ = 0

x² - $\frac{\mathrm{2x}}{2}$ - $\frac{3}{4}$ = 0

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 2 and 4.

The L.C.M is 4.

4 x x² - $\frac{\mathrm{2x}}{2}$ x 4 - 4 x $\frac{3}{4}$ = 0 x 4

4x² - 4x - 3 = 0

When you substitute x = root of the quadratic equation into the equation, the result should be 0.

Since the root is 2, when you substitute x = 2 into 2x² + (k + 2)x + k, the result should be 0.

2x² + (k + 2)x + k

When x = 2,

2(2)² + (k + 2)(2) + k = 0

2 x 4 + 2k + 4 + k = 0

8 + 2k + 4 + k = 0

8 + 4 + 2k + k = 0

12 + 3k = 0

12 = 0 -3k

12 = -3k

Divide both sides by -3.

k = $\frac{12}{-3}$ = -4