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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