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

Answer 3

To solve the given equations:

1. (3x - 2)(x - 3) = 20

Let's first expand the left side of the equation:

3x^2 - 9x - 2x + 6 = 20

Combine like terms:

3x^2 - 11x + 6 = 20

Now, subtract 20 from both sides:

3x^2 - 11x + 6 - 20 = 0

3x^2 - 11x - 14 = 0

Now, we can solve this quadratic equation. You can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 3, b = -11, and c = -14. Plug these values into the quadratic formula:

x = (11 ± √((-11)^2 - 4(3)(-14))) / (2(3))

x = (11 ± √(121 + 168)) / 6

x = (11 ± √289) / 6

x = (11 ± 17) / 6

Now, we have two possible solutions:

1. x = (11 + 17) / 6 = 28 / 6 = 14 / 3
2. x = (11 - 17) / 6 = -6 / 6 = -1

So, the solutions for the first equation are:

x = 14/3 and x = -1

Now, let's solve the second equation:

2. (x + 2)(4x - 5) = -3

Expand the left side:

4x^2 - 5x + 8x - 10 = -3

Combine like terms:

4x^2 + 3x - 10 = -3

Add 3 to both sides:

4x^2 + 3x - 10 + 3 = 0

4x^2 + 3x - 7 = 0

This is another quadratic equation. We can use the quadratic formula again:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4, b = 3, and c = -7. Plug these values into the quadratic formula:

x = (-3 ± √(3^2 - 4(4)(-7))) / (2(4))

x = (-3 ± √(9 + 112)) / 8

x = (-3 ± √121) / 8

x = (-3 ± 11) / 8

Now, we have two possible solutions:

1. x = (-3 + 11) / 8 = 8 / 8 = 1
2. x = (-3 - 11) / 8 = -14 / 8 = -7/4

So, the solutions for the second equation are:

x = 1 and x = -7/4