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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.
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(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.
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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
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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.
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[a + (1/a)]^{2} = 3
Taking square roots both sides, we get
a + (1/a) = √3
On cubing both sides, we
[a + (1/a)]^{3} = (√3)^{3}
[a + (1/a)]^{2} = 3
Taking square roots both sides, we get
a + (1/a) = √3
On cubing both sides, we
[a + (1/a)]^{3} = (√3)^{3}
⇒ a^{3} + 1/a^{3} + 3.a.1/[a(a + 1/a)] = 3 √3
⇒ a^{3} + 1/a^{3} + 3√3 = 3√3
∴ a^{3} + 1/a^{3} = 0