# Quadratic Equation

See how to solve a quadratic equation.

20 examples and their solutions.

- x
^{2}= a

x = ±√a - (x - a)⋅(x - b) = 0

→ x - a = 0

x - b = 0 - x
^{2}+ 8x = 5

x^{2}+ 2⋅x⋅4 + 4^{2}= 5 + 4^{2}

(x + 4)^{2}= 5 + 16 - ax
^{2}+ bx + c = 0 (a ≠ 0)

→ x = -b ± √b^{2}- 4ac2a - D = b
^{2}- 4ac

D > 0, = ■^{2}: 2, rational roots

≠ ■^{2}: 2, irrational roots

D = 0: 1 real root

D < 0: 0 real root - ax
^{2}+ bx + c = 0 (a ≠ 0)

→ Complex roots - x
^{2}- (r_{1}+ r_{2})x + r_{1}r_{2}= 0

ax^{2}+ bx + c = 0 (a ≠ 0)

→ r_{1}+ r_{2}= -ba

r_{1}r_{2}= ca

## Quadratic Equation: Solving by Square Rooting

### Example

x

Solution ^{2}= 9 x

x = ±√9- [1]

= ±3

[1] When square rooting both sides,

write ± in the right side.

Write ± when even number rooting both sides.

(√ ,

^{2}= 9x = ±√9- [1]

= ±3

[1] When square rooting both sides,

write ± in the right side.

Write ± when even number rooting both sides.

(√ ,

^{4}√ ,^{6}√ , ...)### Example

x

Solution ^{2}- 7 = 0 x

x

x = ±√7

^{2}- 7 = 0x

^{2}= 7x = ±√7

### Example

x

Solution ^{2}= -2 x

→ No solution

The left side is x

It's either 0 or (+).

But the right side is (-).

So this quadratic equation has

no solution.

^{2}= -2→ No solution

The left side is x

^{2}.It's either 0 or (+).

But the right side is (-).

So this quadratic equation has

no solution.

## Quadratic Equation: Solving by Factoring

### Example

x

Solution ^{2}- 3x = 0### Example

x

Solution ^{2}+ 5x - 14 = 0 x

(x + 7)(x - 2) = 0

1) x + 7 = 0

x = -7

2) x - 2 = 0

x = 2

x = -7, 2

[1] Factoring

^{2}- 5x - 14 = 0(x + 7)(x - 2) = 0

1) x + 7 = 0

x = -7

2) x - 2 = 0

x = 2

x = -7, 2

[1] Factoring

### Example

x

Solution ^{2}+ 12x + 36 = 0### Example

x

(by factoring)

Solution ^{2}= 9(by factoring)

x

x

(x + 3)(x - 3) = 0- [1]

1) x + 3 = 0

x = -3

2) x - 3 = 0

x = 3

x = ±3

[1] Factoring

^{2}= 9x

^{2}- 3^{2}= 0(x + 3)(x - 3) = 0- [1]

1) x + 3 = 0

x = -3

2) x - 3 = 0

x = 3

x = ±3

[1] Factoring

## Quadratic Equation: Completing the Square

Use the completing the square method

when the x coefficient is even.

### Example

x

Solution ^{2}+ 8x - 5 = 0 x

x

x

(x + 4)

(x + 4)

x + 4 = ±√21

x = -4 ± √21

[1] Use x

to make a perfect square trinomial.

x

→ x

+4

So write +4

[2] Factoring

Close

^{2}+ 8x - 5 = 0x

^{2}+ 8x = 5x

^{2}+ 2⋅x⋅4 + 4^{2}= 5 + 4^{2}- [1](x + 4)

^{2}= 5 + 16- [2](x + 4)

^{2}= 21x + 4 = ±√21

x = -4 ± √21

[1] Use x

^{2}+ 8xto make a perfect square trinomial.

x

^{2}+ 8x→ x

^{2}+ 2⋅x⋅4 + 4^{2}+4

^{2}is added in the left side.So write +4

^{2}in the right side.[2] Factoring

Close

## Quadratic Formula

### Formula

ax

→ x = -b ± √b

Proof ^{2}+ bx + c = 0 (a ≠ 0)→ x = -b ± √b

^{2}- 4ac2a ax

ax

x

x

(x + b2a)

= b

(x + b2a)

x + b2a = ±√b

= ±√b

x = -b2a ± √b

x = -b ± √b

[1] Completing the Square

Close

^{2}+ bx + c = 0ax

^{2}+ bx = -cx

^{2}+ bax = -cax

^{2}+ 2⋅x⋅b2a + (b2a)^{2}= (b2a)^{2}- ca- [1](x + b2a)

^{2}= b^{2}4a^{2}- c⋅4aa⋅4a= b

^{2}4a^{2}- 4ac4a^{2}(x + b2a)

^{2}= b^{2}- 4ac4a^{2}x + b2a = ±√b

^{2}- 4ac4a^{2}= ±√b

^{2}- 4ac2ax = -b2a ± √b

^{2}- 4ac2ax = -b ± √b

^{2}- 4ac2a[1] Completing the Square

Close

### Example

x

Solution ^{2}+ 3x - 2 = 0 x

x = -3 ± √3

= -3 ± √9 + 82

= -3 ± √172

^{2}+ 3x - 2 = 0x = -3 ± √3

^{2}- 4⋅1⋅(-2)2⋅1= -3 ± √9 + 82

= -3 ± √172

### Example

4x

Solution ^{2}- x + 5 = 0 4x

x = -(-1) ± √(-1)

= 1 ± √1 - 808

= 1 ± √-798- [1]

→ No real roots

[1] The number in √ is (-).

Then x is not a real number.

So the quadratic equation has

no real roots.

Complex Roots

Close

^{2}- x + 5 = 0x = -(-1) ± √(-1)

^{2}- 4⋅4⋅52⋅4= 1 ± √1 - 808

= 1 ± √-798- [1]

→ No real roots

[1] The number in √ is (-).

Then x is not a real number.

So the quadratic equation has

no real roots.

Complex Roots

Close

## Discriminant

### Definition

ax

→ x = -b ± √b

→ D = b

The discriminant D,^{2}+ bx + c = 0 (a ≠ 0)→ x = -b ± √b

^{2}- 4ac2a→ D = b

^{2}- 4acis the square root part

of the quadratic formula.

D determines the nature of the roots

of the quadratic equation.

So you can find the nature of the roots

without solving the quadratic equation.

### Nature of the Roots

D > 0, = ■

≠ ■

D = 0: 1 real root

D < 0: 0 real roots

^{2}: 2, rational roots≠ ■

^{2}: 2, irrational rootsD = 0: 1 real root

D < 0: 0 real roots

### Example

x

Nature of the roots?

Solution ^{2}+ 7x + 10 = 0Nature of the roots?

x

D = 7

= 49 - 40

= 9

= 3

Two rational roots- [1]

[1] D = 3

D > 0

Then the quadratic equation has

two rational roots.

Close

^{2}+ 7x + 10 = 0D = 7

^{2}- 4⋅1⋅10= 49 - 40

= 9

= 3

^{2}> 0Two rational roots- [1]

[1] D = 3

^{2}D > 0

Then the quadratic equation has

two rational roots.

Close

### Example

x

Nature of the roots?

Solution ^{2}- 4x - 1 = 0Nature of the roots?

x

D = (-4)

= 16 + 4

= 20 > 0

Two irrational roots- [1]

[1] D = 20 ≠ ■

D > 0

Then the quadratic equation has

two irrational roots.

Close

^{2}- 4x - 1 = 0D = (-4)

^{2}- 4⋅1⋅(-1)= 16 + 4

= 20 > 0

Two irrational roots- [1]

[1] D = 20 ≠ ■

^{2}D > 0

Then the quadratic equation has

two irrational roots.

Close

### Example

x

Nature of the roots?

Solution ^{2}- 6x + 9 = 0Nature of the roots?

x

D = (-6)

= 36 - 36

= 0

One real root

Close

^{2}- 6x + 9 = 0D = (-6)

^{2}- 4⋅1⋅9= 36 - 36

= 0

One real root

Close

### Example

x

Nature of the roots?

Solution ^{2}+ 2x + 5 = 0Nature of the roots?

x

D = 2

= 4 - 20

= -16 < 0

No real roots

Close

^{2}+ 2x + 5 = 0D = 2

^{2}- 4⋅1⋅5= 4 - 20

= -16 < 0

No real roots

Close

## Quadratic Equation: Complex Roots

### Formula

ax

→ x = -b ± √b

If you know what a complex number is,^{2}+ bx + c = 0 (a ≠ 0)→ x = -b ± √b

^{2}- 4ac2ayou can find the complex roots

by using the quadratic formula.

### Example

4x

Solution ^{2}- x + 5 = 0 4x

x = -(-1) ± √(-1)

= 1 ± √1 - 808

= 1 ± √-798

= 1 ± √79i8- [1]

[1] Complex Number

Close

^{2}- x + 5 = 0x = -(-1) ± √(-1)

^{2}- 4⋅4⋅52⋅4= 1 ± √1 - 808

= 1 ± √-798

= 1 ± √79i8- [1]

[1] Complex Number

Close

### Discriminant

D > 0, = ■

≠ ■

D = 0: 1 real root

D < 0: 2 complex roots (= 0 real roots)

If you know what a complex number is,^{2}: 2, rational roots≠ ■

^{2}: 2, irrational rootsD = 0: 1 real root

D < 0: 2 complex roots (= 0 real roots)

you can describe D < 0 differently.

### Example

x

Nature of the roots?

Solution ^{2}+ 2x + 5 = 0Nature of the roots?

x

D = 2

= 4 - 20

= -16 < 0

Two complex roots

Close

^{2}+ 2x + 5 = 0D = 2

^{2}- 4⋅1⋅5= 4 - 20

= -16 < 0

Two complex roots

Close

## Quadratic Equation: Sum and Product of the Roots

### Formula: Roots → Quadratic Equation

x = r

→ x

If the roots of a quadratic equation are_{1}, r_{2}→ x

^{2}- (r_{1}+ r_{2})x + r_{1}r_{2}= 0r

_{1}and r

_{2},

then the quadratic equation is

x

^{2}- (r

_{1}+ r

_{2})x + r

_{1}r

_{2}= 0.

### Example

x = 3, 4

→ Quadratic equation?

Solution → Quadratic equation?

x = 3, 4

→ x

x

→ x

^{2}- (3 + 4)x + 3⋅4 = 0x

^{2}- 7x + 12 = 0### Example

x = 2 + i

→ Quadratic equation?

Solution → Quadratic equation?

x = 2 + i, 2 - i- [1]

→ x

x

x

[1] If x = a + bi, then x = a - bi.

→ x

^{2}- (2 + i + 2 - i)x + (2 + i)⋅(2 - i) = 0x

^{2}- 4x + (4 + 1) = 0x

^{2}- 4x + 5 = 0[1] If x = a + bi, then x = a - bi.

### Formula: Quadratic Equation → Sum and Product of the Roots

ax

→ r

r

From a quadratic equation,^{2}+ bx + c = 0 (a ≠ 0)→ r

_{1}+ r_{2}= -bar

_{1}r_{2}= cayou can directly find

the sum of the roots (r

_{1}+ r

_{2})

and the product of the roots (r

_{1}r

_{2}).

### Example

x

x = 2

Other root?

Solution ^{2}+ 6x + c = 0x = 2

Other root?

x = 2, r- [1]

x

→ r + 2 = -61

r + 2 = -6

r = -8

[1] One root is 2. Set the other root x = r.

x

^{2}+ 6x + c = 0→ r + 2 = -61

r + 2 = -6

r = -8

[1] One root is 2. Set the other root x = r.

### Example

3x

x = 5

Other root?

Solution ^{2}+ bx - 15 = 0x = 5

Other root?

x = 5, r

3x

→ 5⋅r = -153

5r = -5

r = -1

3x

^{2}+ bx - 15 = 0→ 5⋅r = -153

5r = -5

r = -1