Solve for x 2√3x^25x√3=0 Get the answers you need, now! Best answer 2√3x2 ˗ 5x √3 ⇒ 2√3x2 ˗ 2x ˗ 3x √3 ⇒ 2x (√3x ˗ 1) ˗ √3 (√3x ˗ 1) = 0 ⇒ (√3x ˗ 1) or (2x − √3) = 0 ⇒ (√3x ˗ 1) = 0 or (2x − √3) = 0 ⇒ x = 1 3 or x = 3 2 ⇒ x = 1 3 x 3 3 = 3 3 or x = 3 2The given equations is 2√3x 2 5x √3 = 0 Comparing it with ax 2 bx c = 0we get a = 2√3, b = 5 and c = √3 ∴ Discriminant, D = b 2 4ac = (5) 2 4 x 2√3 x √3 = 25 25 = 1 > 0 So, the given equation has real roots Now, √D = √1 = 1 Hence, √3/2 and
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(x+3)(3x-2)(5x+8)^2 0-Solve 4√3x^2 5x 2√3 = 0;X^2 4ax 4a^2 b^2 = 0
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Zeangel4413 zeangel4413 Math Secondary School answered Solve for x 2√3x^25x√3=0 1 See answer zeangel4413 is waiting for your help Add your answer and earn pointsSolve √3x^2 10x 8√3 = 0 Solve √3x^2 2√2x 2√3 = 0;X^2 6x (a^2 2a 8) = 0;
Solve for x √3x 22√2x2√3=0 = 0 Solution Question 5 A two digit number is four times the sum of the digits It is also equal to 3 times the product of digits Find the number Solution Question 6 Solve for x√3x 2 – 2√2x2√3=0 Solution Question 7 Solve for x Solution Short Answer Type Questions II 3 Marks Question 8 Solve for x Find the zeroes of the polynomial f(x) = 2√3x^2 ˗ 5x √3 and verify the relation between its zeroes and coefficients asked Jun 17 in Polynomials by Hailley (Find the roots of the following equations, if they exist, by applying the quadratic formula 2√3x^2 5x √3 = 0
(vi) 2√3x 2 – 5x √3 = 0 Answer (i) x 2 – 4x – 1 = 0 Here, a = 1, b = 4, c = 1 ∴ D = b 2 – 4ac = (4) 2 – 4 × 1 × (1) = 16 4 = ∵ D > 0 Roots are real and distinct (ii) 3x 2 – 2x 1/3 = 0 Here a = 3, b = 2, c = 1/3 ∴ D = b 2 – 4ac = (2) 2 – 4 × 3 × 1/3 = 4 – 4 = 0 ∵ D = 0 ∴ Roots are real and equal (iii) 3x 2 4√3x 4 = 0Solve 4x^2 4bx (a^2 b^2) = 0 Solve x^2 5x (a^2 a 6) = 0;
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