herme519 發問於 科學及數學數學 · 7 年前

methods of solving a quadratic equation:

(1) By factorization

(2) By completing the square

(4) By graphical method

### 1 個解答

• 7 年前
最愛解答

我只做 (1), (2), (3)。

(4)要畫圖再看相交點。

Question 28

(1) By factorization

(2x-5)² = (x-3)(6x-15)

(2x-5)² = 3(x-3)(2x-5)

(2x-5)² - 3(x-3)(2x-5) = 0

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

(2x-5)(2x-5-3x+9) = 0

(2x-5)(-x+4) = 0

(2x-5)(x-4) = 0

x = 5/2 or x = 4

(2) By completing the square

(2x-5)² = (x-3)(6x-15)

4x²-20x+25=6x²-15x-18x+45

4x²-20x+25=6x²-33x+45

2x²-13x+20=0

x²-6.5x+10=0

x²-6.5x+3.25² = -10+3.25²

(x-3.25)² = 0.5625

x - 3.25 = ±0.75

x = 3.25±0.75

x = 2.5 or x = 4

(2x-5)² = (x-3)(6x-15)

4x²-20x+25=6x²-15x-18x+45

4x²-20x+25=6x²-33x+45

2x²-13x+20=0

x = [-(-13)±√[(-13)²-4(2)(20)]]/[2*2]

x = (13±3)/4

x = 10/4 or 16/4

x = 5/2 or 4

Question 30

(1) By factorization

(4-x)² - 5(4-x) - 14 = 0

[(4-x)+2][(4-x)-7] = 0

(6-x)(-3-x) = 0

(x-6)(x+3) = 0

x = 6 or x = -3

(2) By completing the square

(4-x)² - 5(4-x) - 14 = 0

16-8x+x²-20+5x-14 = 0

x²-3x-18 = 0

x²-3x = 18

x²-3x+1.5² = 18+1.5²

(x-1.5)² = 20.25

x-1.5 = ±4.5

x = 1.5±4.5

x = 6 or -3

(4-x)² - 5(4-x) - 14 = 0

16-8x+x²-20+5x-14 = 0

x²-3x-18 = 0

x = {-(-3)±√[(-3)²-4(1)(-18)] }/(2*1)

x = {3±9}/2

x = 12/2 or -6/2

x = 6 or -3

Question 32

(1) By factorization

(x+2)² - 81x² = 0

(x+2)² - (9x)² = 0

(x+2+9x)(x+2-9x) = 0

(10x+2)(-8x+2) = 0

(5x+1)(4x-1) = 0

x = -1/5 or x = 1/4

(2) By completing the square

(x+2)² - 81x² = 0

x² + 4x + 4 - 81x² = 0

80x² - 4x - 4 = 0

20x² - x - 1 = 0

x² - x/20 - 1/20 = 0

x² - x/20 + (1/40)² = 1/20 + (1/40)²

(x - 1/40)² = 81/1600

x - 1/40 = ±9/40

x = 1/40 ± 9/40

x = 10/40 or -8/40

x = 1/4 or -1/5

(x+2)² - 81x² = 0

x² + 4x + 4 - 81x² = 0

80x² - 4x - 4 = 0

20x² - x - 1 = 0

x = [-(-1)±√[(-1)²-4(20)(-1)]]/[2*20]

x = (1±9)/40

x = -8/40 or 10/40

x = -1/5 or 1/4

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