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.
11 and 12 Grade Math
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