Solve the in equation 15 x 2 + 4 x - 4 ≤ 0 by algebraic method.

Given, 15 x 2 + 4 x - 4 ≤ 0 ⇒ 15 x 2 + 10 x - 6 x - 4 ≤ 0 ⇒ ( 3 x + 2 ) ( 5 x - 2 ) ≤ 0 ⇒ x + 2 3 x - 2 5 ≤ 0 ⇒ - 2 3 ≤ x ≤ 2 5

Solve the in equation 15 x 2 + 4 x - 4 ≤ 0 by algebraic method.

Given, 15 x 2 + 4 x - 4 ≤ 0 ⇒ 15 x 2 + 10 x - 6 x - 4 ≤ 0 ⇒ ( 3 x + 2 ) ( 5 x - 2 ) ≤ 0 ⇒ x + 2 3 x - 2 5 ≤ 0 ⇒ - 2 3 ≤ x ≤ 2 5

Solve the in equation x 2 - 2 x + 1 < 0 , x ∈ R by algebraic method.

Given, x 2 - 2 x + 1 < 0 ⇒ ( x - 1 ) 2 < 0 Since, x ∈ R , so ( x - 1 ) 2 ≥ 0 Therefore, no solution.

Solve the in equation 2 - 3 x - 2 x 2 ≥ 0 by algebraic method.

Given, 2 - 3 x - 2 x 2 ≥ 0 Multiplying both sides by - 1 we get 2 x 2 + 3 x - 2 ≤ 0 ⇒ 2 x 2 + 4 x - x - 2 ≤ 0 ⇒ 2 x 2 + 4 x - x - 2 ≤ 0 ⇒ 2 x ( x + 2 ) - 1 ( x + 2 ) ≤ 0 ⇒ ( 2 x - 1 ) ( x + 2 ) ≤ 0 ⇒ - 2 ≤ x ≤ 1 2

Solve the in equation x 2 - 4 x - 21 ≥ 0 by algebraic method.

Given, x 2 - 4 x - 21 ≥ 0 ⇒ x 2 - 7 x + 3 x - 21 ≥ 0 ⇒ x x - 7 + 3 x - 7 ≥ 0 ⇒ x - 7 x + 3 ≥ 0 ⇒ x ⩾ 7 ∪ x ⩽ - 3

Solve the in equation 15 x 2 + 4 x - 4 ≤ 0 by graphical method.

Given, 15 x 2 + 4 x - 4 ≤ 0 ⇒ 15 x 2 + 10 x - 6 x - 4 ≤ 0 ⇒ 5 x ( 3 x + 2 ) - 2 ( 3 x + 2 ) ≤ 0 ⇒ - 2 3 ≤ x ≤ 2 5

Solve the in equation x 2 - 4 x - 21 ≥ 0 by graphical method.

Given, x 2 - 4 x - 21 ≥ 0 Let, y = x 2 - 4 x - 21 ⇒ y = x 2 - 7 x + 3 x - 21 ⇒ y = ( x - 7 ) ( x - 3 ) x 0 1 2 3 4 5 6 7 8 y 21 12 5 0 - 3 - 4 - 3 0 5 From the graph, we can see that y ≥ 0 for x ≤ 3 and x ≥ 7 ∴ Solution of x 2 - 4 x - 21 ≥ 0 is ( - ∞ , 3 ] ∪ [ 7 , ∞ ) .