We will solve different types of problems on quadratic equation using quadratic formula and by method of completing the squares. We know the general form of the quadratic equation i.e., ax$$^{2}$$ + bx + c = 0, that will help us to find the nature of the roots and formation of the quadratic equation whose roots are given.

1. Solve the quadratic equation 3x$$^{2}$$ + 6x + 2 = 0 using quadratic formula.

Solution:

The given quadratic equation is 3x$$^{2}$$ + 6x + 2 = 0.

Now comparing the given quadratic equation with the general form of the quadratic equation ax$$^{2}$$ + bx + c = 0 we get,

a = 3, b = 6 and c = 2

Therefore, x = $$\frac{- b ± \sqrt{b^{2} - 4ac}}{2a}$$

⇒ x = $$\frac{- 6 ± \sqrt{6^{2} - 4(3)(2)}}{2(3)}$$

⇒ x = $$\frac{- 6 ± \sqrt{36 - 24}}{6}$$

⇒ x = $$\frac{- 6 ± \sqrt{12}}{6}$$

⇒ x = $$\frac{- 6 ± 2\sqrt{3}}{6}$$

⇒ x = $$\frac{- 3 ± \sqrt{3}}{3}$$

Hence, the given quadratic equation has two and only two roots.

The roots are $$\frac{- 3 - \sqrt{3}}{3}$$ and $$\frac{- 3 - \sqrt{3}}{3}$$.

2. Solve the equation 2x$$^{2}$$ - 5x + 2 = 0 by the method of completing the squares.

Solutions:

The given quadratic equation is 2x$$^{2}$$ - 5x + 2 = 0

Now dividing both sides by 2 we get,

x$$^{2}$$ - $$\frac{5}{2}$$x + 1 = 0

⇒ x$$^{2}$$ - $$\frac{5}{2}$$x = -1

Now adding $$(\frac{1}{2} \times \frac{-5}{2})$$ = $$\frac{25}{16}$$ on both the sides, we get

⇒ x$$^{2}$$ - $$\frac{5}{2}$$x + $$\frac{25}{16}$$ = -1 + $$\frac{25}{16}$$

⇒ $$(x - \frac{5}{4})^{2}$$ = $$\frac{9}{16}$$

⇒ $$(x - \frac{5}{4})^{2}$$ = ($$\frac{3}{4}$$)$$^{2}$$

⇒ x - $$\frac{5}{4}$$ = ± $$\frac{3}{4}$$

⇒ x = $$\frac{5}{4}$$ ± $$\frac{3}{4}$$

⇒ x = $$\frac{5}{4}$$ - $$\frac{3}{4}$$ and $$\frac{5}{4}$$ + $$\frac{3}{4}$$

⇒ x = $$\frac{2}{4}$$ and $$\frac{8}{4}$$

⇒ x = $$\frac{1}{2}$$ and 2

Therefore, the roots of the given equation are $$\frac{1}{2}$$ and 2.

3. Discuss the nature of the roots of the quadratic equation 4x$$^{2}$$ - 4√3 + 3 = 0.

Solution:

The given quadratic equation is 4x$$^{2}$$ - 4√3 + 3 = 0

Here the coefficients are real.

The discriminant D = b$$^{2}$$ - 4ac = (-4√3 )$$^{2}$$ - 4 4 3 = 48 - 48 = 0

Hence the roots of the given equation are real and equal.

4. The coefficient of x in the equation x$$^{2}$$ + px + q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

Solution:

According to the problem -2 and -15 are the roots of the equation x$$^{2}$$ + 17x + q = 0.

Therefore, the product of the roots = (-2)(-15) = $$\frac{q}{1}$$

⇒ q = 30.

Hence, the original equation is x$$^{2}$$ – 13x + 30 = 0

⇒ (x + 10)(x + 3) = 0

⇒ x = -3, -10

Therefore, the roots of the original equation are -3 and -10.