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โจทย์สมการกำลังสอง แก้โจทย์ด้วยสูตร ควอดราติก

1.	$$x^2 + 4x + 4$$	=	$$0$$
2.	$$x^2 - 4x + 16\sqrt{2}$$	=	$$0$$
3.	$$x^2 + 6x + (3 \times 3)$$	=	$$0$$
4.	$$2x^2 - 8x + 2 + 2 + 4$$	=	$$0$$
5.	$$3x^2 + 6x + 9\sqrt{2}$$	=	$$0$$
6.	$$4x^2 + 8x + 4$$	=	$$0$$
7.	$$4x^2 + 2x + 10x + 9$$	=	$$0$$
8.	$$5x^2 - 10x + 3 + 2$$	=	$$0$$
9.	$$6x^2 + 12x + 36\sqrt{2}$$	=	$$0$$
10.	$$x^2 + 2\sqrt{2}x + 2$$	=	$$0$$

โจทย์สมการกำลังสอง : ข้อที่ 1

$$x^2 + 4x + 4$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$4^2 - (4 \times 1 \times 4)$$
$$D$$	=	$$16 - (16)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-4 \pm \sqrt{4^2 - (4 \times 1 \times 4)} \over (2 \times 1)$$
$$x$$	=	$$-4 \pm \sqrt{4^2 - (16)} \over 2$$
$$x$$	=	$$-4 \pm \sqrt{16 - (16)} \over 2$$
$$x$$	=	$$-4 \pm \sqrt{0} \over 2$$
$$x$$	=	$$-4 \pm 0 \over 2$$
$$x$$	=	$$-4 \over 2$$
$$x$$	=	$$-2$$

โจทย์สมการกำลังสอง : ข้อที่ 2

$$x^2 - 4x + 16\sqrt{2}$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$4^2 - (4 \times 1 \times (16 \sqrt 2))$$
$$D$$	=	$$4^2 - (4 \times 1 \times (4))$$
$$D$$	=	$$16 - (16)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-4 \pm \sqrt{4^2 - (4 \times 1 \times 4)} \over (2 \times 1)$$
$$x$$	=	$$-4 \pm \sqrt{4^2 - (16)} \over 2$$
$$x$$	=	$$-4 \pm \sqrt{16 - (16)} \over 2$$
$$x$$	=	$$-4 \pm \sqrt{0} \over 2$$
$$x$$	=	$$-4 \pm 0 \over 2$$
$$x$$	=	$$-4 \over 2$$
$$x$$	=	$$-2$$

โจทย์สมการกำลังสอง : ข้อที่ 3

$$x^2 + 6x + (3 \times 3)$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$6^2 - (4 \times 1 \times (3 \times 3))$$
$$D$$	=	$$6^2 - (4 \times 1 \times (9))$$
$$D$$	=	$$6^2 - (36)$$
$$D$$	=	$$36 - (36)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-6 \pm \sqrt{6^2 - (4 \times 1 \times (3 \times 3))} \over (2 \times 1)$$
$$x$$	=	$$-6 \pm \sqrt{6^2 - (4 \times 1 \times (9))} \over (2)$$
$$x$$	=	$$-6 \pm \sqrt{6^2 - (4 \times 1 \times 9)} \over 2$$
$$x$$	=	$$-6 \pm \sqrt{6^2 - 36} \over 2$$
$$x$$	=	$$-6 \pm \sqrt{36 - 36} \over 2$$
$$x$$	=	$$-6 \pm \sqrt{0} \over 2$$
$$x$$	=	$$-6 \over 2$$
$$x$$	=	$$-3$$

โจทย์สมการกำลังสอง : ข้อที่ 4

$$2x^2 - 8x + 2 + 2 + 4$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$(-8)^2 - (4 \times 2 \times (2 + 2 + 4))$$
$$D$$	=	$$(-8)^2 - (4 \times 2 \times (8))$$
$$D$$	=	$$(-8)^2 - (64)$$
$$D$$	=	$$64 - (64)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(-8) \pm \sqrt{(-8)^2 - (4 \times 2 \times (2 + 2 + 4))} \over (2 \times 2)$$
$$x$$	=	$$8 \pm \sqrt{(-8)^2 - (4 \times 2 \times (2 + 2 + 4))} \over (2 \times 2)$$
$$x$$	=	$$8 \pm \sqrt{(-8)^2 - (4 \times 2 \times (2 + 2 + 4))} \over (4)$$
$$x$$	=	$$8 \pm \sqrt{(-8)^2 - (4 \times 2 \times (8))} \over (4)$$
$$x$$	=	$$8 \pm \sqrt{(-8)^2 - (64)} \over 4$$
$$x$$	=	$$8 \pm \sqrt{64 - (64)} \over 4$$
$$x$$	=	$$8 \pm \sqrt{0} \over 4$$
$$x$$	=	$$8 \over 4$$
$$x$$	=	$$2$$

โจทย์สมการกำลังสอง : ข้อที่ 5

$$3x^2 + 6x + 9\sqrt{2}$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$6^2 - (4 \times 3 \times (9\sqrt{2}))$$
$$D$$	=	$$6^2 - (4 \times 3 \times (3))$$
$$D$$	=	$$6^2 - (36)$$
$$D$$	=	$$36 - (36)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(-6) \pm \sqrt{(-6)^2 - (4 \times 3 \times (9\sqrt{2}))} \over (2 \times 3)$$
$$x$$	=	$$6 \pm \sqrt{(-6)^2 - (4 \times 3 \times (9\sqrt{2}))} \over (2 \times 3)$$
$$x$$	=	$$6 \pm \sqrt{(-6)^2 - (4 \times 3 \times (9\sqrt{2}))} \over (6)$$
$$x$$	=	$$6 \pm \sqrt{(-6)^2 - (4 \times 3 \times (3))} \over (6)$$
$$x$$	=	$$6 \pm \sqrt{(-6)^2 - (36)} \over 6$$
$$x$$	=	$$6 \pm \sqrt{36 - (36)} \over 6$$
$$x$$	=	$$6 \pm \sqrt{0} \over 6$$
$$x$$	=	$$6 \over 6$$
$$x$$	=	$$1$$

โจทย์สมการกำลังสอง : ข้อที่ 6

$$4x^2 + 8x + 4$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$8^2 - (4 \times 4 \times 4)$$
$$D$$	=	$$8^2 - (64)$$
$$D$$	=	$$64 - (64)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(8) \pm \sqrt{8^2 - (4 \times 4 \times 4)} \over (2 \times 4)$$
$$x$$	=	$$-(8) \pm \sqrt{8^2 - (4 \times 4 \times 4)} \over (8)$$
$$x$$	=	$$-(8) \pm \sqrt{8^2 - (64)} \over 8$$
$$x$$	=	$$-(8) \pm \sqrt{64 - (64)} \over 8$$
$$x$$	=	$$-(8) \pm \sqrt{0} \over 8$$
$$x$$	=	$$-(8) \over 8$$
$$x$$	=	$$-1$$

โจทย์สมการกำลังสอง : ข้อที่ 7

$$4x^2 + 2x + 10x + 9$$	=	$$0$$
$$4x^2 + 12x + 9$$	=	$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$12^2 - (4 \times 4 \times 9)$$
$$D$$	=	$$12^2 - (144)$$
$$D$$	=	$$144 - (144)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (4 \times 4 \times 9)} \over (2 \times 4)$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (4 \times 4 \times 9)} \over (8)$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (144)} \over 8$$
$$x$$	=	$$-(12) \pm \sqrt{144 - (144)} \over 8$$
$$x$$	=	$$-(12) \pm \sqrt{0} \over 8$$
$$x$$	=	$$-(12) \over 8$$
$$x$$	=	$$-(3) \over 2$$
$$x$$	=	$$-{3 \over 2}$$

โจทย์สมการกำลังสอง : ข้อที่ 8

$$5x^2 - 10x + 3 + 2$$	=	$$0$$
$$5x^2 - 10x + 5$$	=	$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$(-10)^2 - (4 \times 5 \times 5)$$
$$D$$	=	$$(-10)^2 - (100)$$
$$D$$	=	$$100 - (100)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(-10) \pm \sqrt{(-10)^2 - (4 \times 5 \times 5)} \over (2 \times 5)$$
$$x$$	=	$$10 \pm \sqrt{(-10)^2 - (4 \times 5 \times 5)} \over (2 \times 5)$$
$$x$$	=	$$10 \pm \sqrt{(-10)^2 - (4 \times 5 \times 5)} \over (10)$$
$$x$$	=	$$10 \pm \sqrt{(-10)^2 - (100)} \over 10$$
$$x$$	=	$$10 \pm \sqrt{100 - (100)} \over 10$$
$$x$$	=	$$10 \pm \sqrt{0} \over 10$$
$$x$$	=	$$10 \over 10$$
$$x$$	=	$$1$$

โจทย์สมการกำลังสอง : ข้อที่ 9

$$6x^2 + 12x + 36\sqrt{2}$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$12^2 - (4 \times 6 \times (36\sqrt{2}))$$
$$D$$	=	$$12^2 - (4 \times 6 \times (6))$$
$$D$$	=	$$12^2 - (144)$$
$$D$$	=	$$144 - (144)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (4 \times 6 \times (36\sqrt{2}))} \over (2 \times 6)$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (4 \times 6 \times (36\sqrt{2}))} \over (12)$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (4 \times 6 \times (6))} \over (12)$$
$$x$$	=	$$-(12) \pm \sqrt{12^2 - (144)} \over 12$$
$$x$$	=	$$-(12) \pm \sqrt{144 - (144)} \over 12$$
$$x$$	=	$$-(12) \pm \sqrt{0} \over 12$$
$$x$$	=	$$-(12) \over 12$$
$$x$$	=	$$-1$$

โจทย์สมการกำลังสอง : ข้อที่ 10

$$x^2 + 2\sqrt{2}x + 2$$

=

$$0$$

$$D$$	=	$$b^2 - 4ac$$
$$D$$	=	$$(2\sqrt{2})^2 - ({4}\times{1}\times{2})$$
$$D$$	=	$$8 - (8)$$
$$D$$	=	$$0$$ ดังนั้น มีหนึ่งคำตอบ

Quadratic Equation	=	$$-b \pm \sqrt{b^2 - 4ac} \over 2a$$
$$x$$	=	$$-(2\sqrt{2}) \pm \sqrt{(2\sqrt{2})^2 - (4 \times 1 \times 2)} \over 2 \times 1$$
$$x$$	=	$$-(2\sqrt{2}) \pm \sqrt{(2\sqrt{2})^2 - 8} \over 2$$
$$x$$	=	$$-(2\sqrt{2}) \pm \sqrt{8 - 8} \over 2$$
$$x$$	=	$$-(2\sqrt{2}) \pm \sqrt{0} \over 2$$
$$x$$	=	$$-(2\sqrt{2}) \pm 0 \over 2$$
$$x$$	=	$$-2\sqrt{2} \over 2$$
$$x$$	=	$$-\sqrt{2}$$