Question

find the roots of quadratic equation by using the quadratic formula in each of the following: (1) x2-3√5x+10=0 ​

Answer 1

Answer :

2√5 and √5

Step-by-step explanation :

    Quadratic Polynomials :

        ✯ It is a polynomial of degree 2

        ✯ General form :

                  ax² + bx + c  = 0

                 

        ✯ Quadratic formula :

               $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

        ✯ Determinant, D = b² - 4ac

        ✯ Based on the value of Determinant, we can define the nature of roots.

                D > 0 ; real and unequal roots

                D = 0 ; real and equal roots

                D < 0 ; no real roots i.e., imaginary

        ✯ Relationship between zeroes and coefficients :

                  ✩ Sum of zeroes = -b/a

                  ✩ Product of zeroes = c/a

___________________________

              Given quadratic equation,

                 x² - 3√5x + 10 = 0

 =>   It is of the form , ax² + bx +c = 0

         a = 1, b = -3√5, c = 10

we know,

        $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

  substitute the values of a,b,c

        $x=\frac{-(-3\sqrt{5})\pm\sqrt{(-3\sqrt{5})^2-4(1)(10)} }{2(1)} \\\\ x=\frac{3\sqrt{5}\pm\sqrt{9(5)-40} }{2} \\\\x=\frac{3\sqrt{5}\pm\sqrt{45-40}}{2} \\\\ x=\frac{3\sqrt{5}\pm\sqrt{5}}{2} \\\\ x=\frac{\sqrt{5}(3\pm1)}{2} \\\\ x=\frac{\sqrt{5}(3+1)}{2} \ \ || \ \ x=\frac{\sqrt{5}(3-1)}{2} \\\\ x=\frac{\sqrt{5}(4)}{2} \ \ \ \ \ || \ \ x=\frac{\sqrt{5}(2)}{2} \\\\ x=2\sqrt{5} \ \ \ \ \ \ || \ \ x=\sqrt{5}$

∴ The roots of the given equation are 2√5 and √5