Worksheet on Quadratic Formula

Practice the questions given in the worksheet on quadratic formula. We know the solutions of the general form of the quadratic equation ax\(^{2}\) + bx + c = 0 are x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).

1. Answer the following:

(i) Is it possible to apply quadratic formula in the equation 2t\(^{2}\) +(4t - 1)(4t + 1) = 2t(9t - 1)

(ii) What type of equations can be solved using quadratic formula?

(iii) Applying quadratic formula, solve the equation (z - 2)(z + 4) = - 9

(iv) Applying quadratic formula in the equation 5y\(^{2}\) + 2y - 7 = 0, we get y = \(\frac{k ± 12}{10}\), What is the value of K?

(v) Applying quadratic formula in a quadratic equation, we get m = \(\frac{9 \pm \sqrt{(-9)^{2} - 4 ∙ 14 ∙ 1}}{2 ∙ 14}\). Write the equation.

 

2. With the help of quadratic formula, solve each of the following equations:

(i) x\(^{2}\) - 6x = 27

(ii) \(\frac{4}{x}\) - 3 = \(\frac{5}{2x + 3}\)

(iii) (4x - 3)\(^{2}\) - 2(x + 3) = 0

(iv) x\(^{2}\) - 10x + 21 = 0

(v) (2x + 7)(3x - 8) + 52 = 0

(vi) \(\frac{2x + 3}{x + 3}\) = \(\frac{x + 4}{x + 2}\)

(vii) x\(^{2}\) + 6x - 10 = 0

(viii) (3x + 4)\(^{2}\) - 3(x + 2) = 0

(ix) √6x\(^{2}\) - 4x - 2 √6 = 0

(x) (4x - 2)\(^{2}\) + 6x - 25 = 0

(xi) \(\frac{x - 1}{x - 2}\) + \(\frac{x - 3}{x - 4}\) = 3\(\frac{1}{3}\)

(xii) \(\frac{2x}{x - 4}\) + \(\frac{2x - 5}{x - 3}\) = 8\(\frac{1}{3}\)


Answers for the worksheet on quadratic formula are given below.


Answers:


1. (i) No

(ii) Quadratic equation in one variable

(iii) -1, -1

(iv) K = -2

(v) 14m\(^{2}\) - 9m + 1 = 0

 

2. (i) -3 or 9

(ii) -2 or 1

(iii) x = \(\frac{3}{2}\) or \(\frac{1}{8}\)

(iv) 3 or 7

(v) x = -\(\frac{4}{3}\) or \(\frac{1}{2}\)

(vi) ±√6

(vii) -3 ± √19

(viii) x = -\(\frac{5}{3}\) or -\(\frac{2}{3}\)

(ix) √6 or -\(\frac{√6 }{3}\)

(x) x = -\(\frac{7}{8}\) or \(\frac{3}{2}\)

(xi) 2\(\frac{1}{2}\) or 5

(xii) 3\(\frac{1}{13}\) or 6





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