Question

5.25. a) 6) (3x - 2) (x - 3) = 20; (x + 2) (4x - 5) = -3;

Answer 1

Ответ:

a) $x=-1, \ x= \frac{ 14 }{ 3 }$

б) $x=- \frac{ 7 }{ 4 },\ x=1$

Объяснение:

a)

$(3x - 2) (x - 3) = 20\\ 3x^2-9x-2x+6-20=0\\3 x^2-11x-14=0\\\\a=3 ,\ \ b=-11 ,\ \ c=-14$

$\ D = b^2 - 4ac = ( - 11)^2 - 4\cdot3\cdot( - 14) = 121 + 168 = 289\\\\\\\sqrt{D} =\sqrt{289} = 17\\\\\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{11-17}{2\cdot3}=\frac{-6 }{6 }=-1\\\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{11+17}{2\cdot3}=\frac{28}{6}= \frac{ 14 }{ 3 }$

б)

$(x + 2) (4x - 5) =- 3\\ 4x^2-5x+8x-10+3=0\\ 4x^2+3x-7=0\\a=4 ,\ \ b=3 ,\ \ c=-7$

$\ D = b^2 - 4ac = 3^2 - 4\cdot4\cdot(- 7) = 9 + 112 = 121\\\\\\\sqrt{D} =\sqrt{121} = 11\\\\\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-3-11}{2\cdot4}=\frac{-14 }{8 }=- \frac{ 7 }{ 4 } \\\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-3+11}{2\cdot4}=\frac{8}{8}=1$

Answer 2

Объяснение:

а)

(3х - 2)(х - 3) = 20

3х² - 9х - 2х + 6 - 20 = 0

3х² - 11х - 14 = 0

а = 3; в = -11; с = -14

Д = в² - 4ас

Д = (-11)² - 4 * 3 * (-14) = 121 + 168 = 289

√Д = √289 = 17

х1 = (-в - √Д)/2а

х1 = (11 - 17)/(2*3) = (-6)/6 = -1

х2 = (-в + √Д)/2а

х2 = (11 + 17)/(2*3) = 28/6 = 14/3 = 4 2/3

б)

(х + 2)(4х - 5) = -3

4х² - 5х + 8х - 10 + 3 = 0

4х² + 3х - 7 = 0

а = 4; в = 3; с = -7

Д = в² - 4ас

Д = 3² - 4 * 4 * (-7) = 9 + 112 = 121

√Д = √121 = 11

х1 = (-в - √Д)/2а

х1 = (-3 - 11)/(2*4) = -14/8 = -7/4 = -1 3/4

х2 = (-в + √Д)/2а

х2 = (-3 + 11)/(2*4) = 8/8 = 1