Show your solution follow the steps need an answer, please need an answer, please.

Step-by-step explanation:
21. x² - 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
m² - 11m + 18 = 0
hence, x + y = 11
xy = 18
2xy = 36
now, x² + y²
= (x +y)² - 2xy
= (11)² - 36
= 121 - 36
= 85
therefore, option A is correct.
26. x² - 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)