-- advertisement --
View Hint | View Answer | Workspace | Discuss In Forum | Report
Quadratic equation must be in the form of ax^{2} + bx + c = 0, where a ≠ 0.
Clearly, 7x^{2} = 49 or 7x^{2} - 49 = 0, which is of the form ax^{2} + bx + c = 0, where b = 0.
Thus, 7x^{2} - 49 = 0 is a quadratic equation.
View Hint | View Answer | Workspace | Discuss In Forum | Report
(x - 1) (2x - 5) = 0 ⇒ x = 1, 5/2
So, its roots are real.
(x - 1) (2x - 5) = 0 ⇒ x = 1, 5/2
So, its roots are real.
View Hint | View Answer | Workspace | Discuss In Forum | Report
Given equation is 2x^{2} - 9x - 18 = 0
[by factorisation method]
⇒ 2x^{2} - 12x + 3x - 18 = 0
Given equation is 2x^{2} - 9x - 18 = 0
[by factorisation method]
⇒ 2x^{2} - 12x + 3x - 18 = 0
⇒ 2x(x - 6) + 3(x - 6) = 0
⇒ (2x + 3) (x - 6) = 0
∴ x = -3/2, 6
View Hint | View Answer | Workspace | Discuss In Forum | Report
Given, one root of x^{2} - 6kx + 5 = 0 is 5.
∴ x = 5 satisfies x^{2} - 6kx + 5 = 0
Given, one root of x^{2} - 6kx + 5 = 0 is 5.
∴ x = 5 satisfies x^{2} - 6kx + 5 = 0
⇒ 5^{2} - 30k + 5 = 0
⇒ 25 - 30k + 5 = 0
⇒ 30 - 30k = 0
⇒ 30k = 30
∴ k = 1
View Hint | View Answer | Workspace | Discuss In Forum | Report
2x^{2} -11x + 15 = 0
[by factorisation method]
⇒ 2x^{2} - (6x + 5x) + 15 = 0
2x^{2} -11x + 15 = 0
[by factorisation method]
⇒ 2x^{2} - (6x + 5x) + 15 = 0
⇒ 2x^{2} - 6x - 5x + 15 = 0
⇒ 2x(x - 3) - 5 (x - 3) = 0
⇒ (2x - 5) (x - 3) = 0
∴ x = 5/2, 3
Hence, the roots are 5/2 and 3.
-- advertisement --