Aryabhata's Contributions to Algebra

For equal roots, the discriminant (b^2 - 4ac) must be equal to zero. Substituting (a) with (\frac{-b}{2a}) in the equation (ax^2 + bx + c = 0) satisfies this condition.

The formula for the sum of the first (n) natural numbers is (\frac{n(n+1)}{2}).

Using the quadratic formula, (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), we can solve for (x) in the equation (x^2 - 2x + 1 = 0). Substituting the values of (a), (b), and (c), we get (x = 1).

The formula for finding the product of two binomials ((a + b)) and ((c + d)) is ((a + b)(c + d) = ac + ad + bc + bd).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^3 - 3x^2 + 3x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^2 - 2x + 1). Solving for (x) in this quadratic equation, we get (x = 1).

The formula for finding the area of a triangle with base (b) and height (h) is (\frac{1}{2}bh).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^4 - 2x^3 + x^2 - 2x + 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^3 - x^2 + 1). Solving for (x) in this cubic equation, we get (x = 1).

The formula for finding the volume of a cube with side (a) is (a^3).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^5 - 3x^4 + 3x^3 - x^2 + x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^4 - 2x^3 + 2x^2 - x + 1). Solving for (x) in this quartic equation, we get (x = 1).

The formula for finding the surface area of a cube with side (a) is (6a^2).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^6 - 4x^5 + 6x^4 - 4x^3 + x^2 - 4x + 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^5 - 3x^4 + 3x^3 - x^2 + x - 1). Solving for (x) in this quintic equation, we get (x = 1).

The formula for finding the volume of a sphere with radius (r) is (\frac{4}{3}\pi r^3).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^7 - 5x^6 + 10x^5 - 10x^4 + 5x^3 - x^2 + x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^6 - 4x^5 + 6x^4 - 4x^3 + x^2 - 4x + 1). Solving for (x) in this sextic equation, we get (x = 1).

The formula for finding the surface area of a sphere with radius (r) is (4\pi r^2).