Case Study I

Shridharacharya was an Indian mathematician, Sanskrit Pandit and philosopher from Bengal. He is known for his treatises – Trisatika and Patiganita. He was the first to give an algorithm for solving quadratic equations. His other major works were on algebra, particularly fractions and he gave an exposition on zero. He separated Algebra from Arithmetic. The quadratic formula which is used to find the roots of a quadratic equation is known as Shridharacharya’s rule.

1. A quadratic equation of the form ax² + bx + c = 0, a ≠ 0, has two roots given by:

1. $x = \dfrac{b \pm \sqrt{b^{2} - 4abc}}{2a}$

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

3. $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2ac}$

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

2. A quadratic equation of the form ax² + bx + c = 0, a ≠ 0, has rational roots, if the value of (b² - 4ac) is:

1. less than 0
2. greater than 0
3. equal to 0
4. equal to 0 or a perfect square

3. The maximum number of roots that a quadratic equation can have is:

1. 1
2. 2
3. 3
4. 4

4. A quadratic equation ax² + bx + c = 0, a ≠ 0, having real coefficients, cannot have real roots if:

1. b² - 4ac < 0
2. b² - 4ac = 0
3. b² - 4ac > 0
4. b² - 4ac ≥ 0

5. The quadratic equation, x² + 26x + 169 = 0, has:

1. non real roots
2. rational and unequal roots
3. equal roots
4. irrational roots

1. By formula,

$x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Hence, option (d) is the correct option.

2. Given,

ax² + bx + c = 0, a ≠ 0

We know that,

For the roots to be rational, the discriminant must be a perfect square or zero.

Hence, option (d) is the correct option.

3. A polynomial of degree n has at most n roots.

Given, equation :

ax² + bx + c = 0

Here, the highest power of x is 2. Therefore, maximum number of roots for quadratic equation is 2.

Hence, option (b) is the correct option.

4. Given,

A quadratic equation cannot have real roots if discriminant is less than zero.

Thus, b² − 4ac < 0.

Hence, option (a) is the correct option.

5. Given,

x² + 26x + 169 = 0

Comparing x² + 26x + 169 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 26 and c = 169.

We know that,

⇒ D = b² - 4ac

⇒ D = (26)² - 4(1)(169)

⇒ D = 676 - 676

⇒ D = 0, the roots are real and equal.

Hence, option (c) is the correct option.