Help me with unsolveable problems?

To denote powers use ^, i.e. x squared = x^2, x cubed = x^3, etc...

2x^2 - 3x - 35 = 0

2x^2 - 10x + 7x - 35 = 0

2x (x - 5) + 7 (x - 5)

(2x + 7) (x - 5) = 0

x = -7/2

x = 5

It looks like you got the 2 coefficient in the wrong factor

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2.

5x(x + 9) = -65

x (x + 9) = -13

x^2 + 9x + 13 = 0

x = (-9 ± √(81 - 4(1)(13))) / 2

x = (-9 ± √(81-52)) / 2

x = (-9 ± √29) / 2

Smaller value: (-9 - √29)/2

Larger value: (-9 + √29)/2

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3.

I'm not exactly sure what sort of example you expected.

Have you never been showed how to cross-multiply, to get rid of denominators?

(x - 2) = 8/x

x (x - 2) = 8

x^2 - 2x - 8 = 0

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

x = -2

x = 4

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4.

This time instead of cross-multiplying, we'll just multiply both sides by least common denominator

1/(x-1) + 1/(x-5) = 3/8

Multiply both sides by lcd = 8(x-1)(x-5)

8(x-1)(x-5)/(x-1) + 8(x-1)(x-5)/(x-5) = 8(x-1)(x-5) * 3/8

8(x-5) + 8(x-1) = 3(x-1)(x-5)

8x - 40 + 8x - 8 = 3 (x^2 - 6x + 5)

16x - 48 = 3x^2 - 18x + 15

0 = 3x^2 - 18x + 15 - 16x + 48

3x^2 - 34x + 63 = 0

3x^2 - 27x - 7x + 63 = 0

3x (x - 9) - 7 (x - 9) = 0

(3x - 7) (x - 9) = 0

x = 7/3

x = 9

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5.

29/b - 23 = -22/(b+1)

Multiply both sides by b(b+1)

29(b+1) - 23b(b+1) = -22b

29b + 29 - 23b^2 - 23b = -22b

-23b^2 + 6b + 29 = -22b

-23b^2 + 28b + 29 = 0

23b^2 - 28b - 29 = 0

b = (28 ± √(784 - 4(23)(-29))) / 46

b = (28 ± √(784 + 2668)) / 46

b = (28 ± √3452) / 46

b = (28 ± 2√863) / 46

b = (14 − √863) / 23

b = (14 + √863) / 23

Mαthmφm

1) 2x²-3x-35 = 0

(2x+7)(x-5) = 0

x = -7/2 = smaller value, 5 = larger value

2) 5x(x+9) = -65

5x²+45x = -65

5x²+45x+65=0

x²+9x+13 = 0

x = (-9±√29)/2 = (-9-√29)/2 = smaller value, (-9+√29)/2 = larger value

3) x-2 = 8/x

x(x-2) = 8

x²-2x-8 = 0

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

x = 4 = larger value, -2 = smaller value

4) 1/(x-1) + 1/(x-5) = 3/8

[(x-5)+(x-1)]/(x-1)(x-5) = 3/8

(2x-6)/(x-1)(x-5) = 3/8

8(2x-6) = 3(x-1)(x-5)

16x-48 = 3(x²-6x+5)

16x-48 = 3x²-18x+15

0 = 3x²-34x+63

0 = (3x-7)(x-9)

x = 7/3 = smaller value, 9 = larger value

5) 29/b - 23 = -22/(b+1)

(29-23b)/b = -22/(b+1)

(29-23b)(b+1) = -22b

6b-23b²-29 = -22b

-23b²+28b-29 = 0

23b²-28b+29 = 0

x = (28±i√1884)/46 = (28-i√1884)/46 = smaller value, (28+i√1884)/46 = larger value

Q1

2x^2 -3x -35=0

(2x+7)(x-5)=0 ........... (check you get factorisation correct by multiplying it back out!)

x=-7/2 or x=5

Q2

to use formula must be in form

ax^2 + bx+ c

5x^2 +45x +65 =0

a=5, b=45 c=65

gives -9 plusminus root29 all over 2

Q3

multiply everything by x to remove the "over x"

x(x-2)=8

x^2 -2x -8=0

then solve as normal

Q4

as above but first multiply by (x-1) and then by (x-5). Simplify as needed, rearrange as solve as normal

Q5

similar but replaced the x's with b's

3/

x(x-2)=8

x^2 - 2x - 8 = 0

(x-4)(x+2)=0 ---> x = 4 or -2

4/

LCD = (x-1)(x-5)

8(x-5)+8(x-1) = 3(x-1)(x-5)

expand , simplify then solve for x

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