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.

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