Q. 204.0( 4 Votes )

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Answer :

Given equation: 2x^{2} + √3 x – 3 = 0

_{Comparing above equation with ax}^{2} + bx + c = 0,

_{⇒} _{a = 2; b = √3; c = -3}

_{Now, b}^{2} – 4ac = (√3)^{2} – 4(2)(-3)

_{= 3 – (-24)}

_{⇒} _{b}^{2} – 4ac = 27 ≥ 0

_{We know that, if b}^{2} – 4ac ≥ 0 then the roots of the quadratic equation ax^{2} + bx + c = 0 is given by _{[–b} _{(√b}^{2}– 4ac)] / 2a which is known as the quadratic formula.

_{⇒} _{x = [-√3} _{√27]/ 2(2)}

_{⇒} _{x = [-√3} _{3√3] / 4}

_{⇒} _{x = [-√3 + 3√3] / 4 (or) x = [-√3 - 3√3] / 4}

⇒ x = (2√3)/4 (or) x = (-4√3)/4

⇒ x = √3/2 (or) x = -√3

Ans. The solutions of x are √3/2 (or) x = -√3.

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