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 = $$\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$.

(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?

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$$\frac{1}{3}$$

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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