Respuesta :

Step-by-step explanation:

21. - 4x - 5 > 0

(x - 5) (x - 1) > 0

{x - 5 > 0

{x + 1 > 0

{x - 5 < 0

{x + 1 < 0

{x > 5

{x > -1

{x < 5

{x < -1

= x < -1 or x > 5

therefore, option A is correct

22). 4p² - 1 0

(2p - 1) (2p + 1) 0

{2p - 1 0

{2p + 1 0

{2p - 1 0

{2p + 1 0

{p 1/2

{p -1/2

{p 1/2

{p -1/2

1/2 p 1/2

{p 1/2

{p -1/2

-1/2 p 1/2

therefore, option A is correct.

23). y < 3x² + 3x - 6

3x² + 3x - 6 = 0

3x² + 6x - 3x - 6 = 0

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

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

x = -2, 1

this is the correct ans. but their is no option similar to it.

25. x and y are root of

- 11m + 18 = 0

hence, x + y = 11

xy = 18

2xy = 36

now, +

= (x +y)² - 2xy

= (11)² - 36

= 121 - 36

= 85

therefore, option A is correct.

26. - 2x - 120

(x -1)² - 1 - 1² = 0

(x -1)² = 1 + 1²

x - 1² = 2

x - 1 = ±2

1 ± 2

therefore, option A is correct.

hope this answer helps you dear..take care!

Answer:

Step-by-step explanation:

21) x² - 4x - 5 > 0

x² - x + 5x - 5 > 0

x(x - 1) + 5(x - 1) > 0

(x - 1) (x + 5) > 0

x - 1 >0   or  x + 5 > 0

x > 1       or  x > - 5    Ans: D

22)  4p² ≤ 1

p² ≤ 1/4

p ≤ 1/2  or  p ≥ -1/2

-1/2  ≥ p ≤ 1/2

23)y < 3x² + 3x - 6

(1 , -7)

x = 1  and y = -7

-7 < 3*1 + 3*1 - 6

-7 < 3 + 3 - 6

-7 < 0

(1 , -7) satisfies the equation.

Option b

24) 5x² - 7x = 0

a = 5  ; b = -7  ;c = 0

Discriminant = b² - 4ac =  49 - 0 = 0

[tex]x = \dfrac{-b+\sqrt{D}}{2a} \ ; \ \dfrac{-b-\sqrt{D}}{2a}\\\\x = \dfrac{7+\sqrt{49}}{10} ; \dfrac{7-\sqrt{49}}{10}\\\\x = \dfrac{7+7}{10};\dfrac{7-7}{10}\\\\x =\dfrac{14}{10} ; 0\\\\x = \dfrac{7}{5}; 0[/tex]

x = 0 ; x = 7/5  Option a)

25) m² - 11 m + 18 = 0

m² - 2m - 9m + (-9)*(-2) = 0

m(m - 2) - 9(m - 2) = 0

(m - 2)(m - 9) = 0

m = 2 , 9

x = 2 and y =9

x² + y² = 2² + 9² = 4 + 81 = 85   Option a)

26) x² - 2x - 1 = 0

Add the constant 1 to both sides

x² - 2x = 1

Divide the coefficient of x by 2.  2/2 = 1. Now add this 1 to both sides of the equation.

x² - 2x  + 1  = 1+1

x² - 2x + 1 = 2

(x - 1)² = 2

Take square root

x - 1 = ±√2

x  = 1 ± √2    Option a)