Question

find zero of the quadratic polynomial 4x^2+5√2x-3​

Answer 1

Step-by-step explanation:

x=1÷root2

or

x=-3÷2root 2

Answer 2

Solution :-

We have to find out the zeros of the Quadratic polynomial :-

4x² + 5√2 x - 3 = 0

Now we will be first converting it into standard Quadratic equation :-

ax² + bx + c where a = 1

In order to do so we will divide it by 4

$\dfrac{4x^2 - 5\sqrt{2}x - 3}{4}$

$= x^2 - \dfrac{5\sqrt{2}}{4} - \dfrac{3}{4}$

Now by Sridhhacharyas formula

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

$x = \dfrac{ \dfrac{5\sqrt{2}}{4} \pm \sqrt{ \left(\dfrac{5\sqrt{2}}{4}\right)^2 - 4 (1)\left(\dfrac{-3}{4}\right) }}{2}$

$x = \dfrac{\dfrac{5\sqrt{2}}{4} \pm \sqrt{\dfrac{25\times 2}{16} + 12 }}{2}$

$x = \dfrac{ \dfrac{5\sqrt{2}}{4} \pm \sqrt{\dfrac{50 + 192}{16}}}{2}$

$x = \dfrac{\dfrac{5\sqrt{2}}{4} \pm \sqrt{\dfrac{242}{16}}}{2}$

$x = \dfrac{\dfrac{5\sqrt{2}}{4} \pm \dfrac{11\sqrt{2}}{4}}{2}$

$x = \dfrac{\dfrac{5\sqrt{2} + 11\sqrt{2}}{4}}{2} \: or \: \dfrac{\dfrac{5\sqrt{2} - 11\sqrt{2}}{4}}{2}$

$x = \dfrac{\dfrac{16\sqrt{2}}{4}}{2} \: or \: \dfrac{\dfrac{-6\sqrt{2}}{4}}{2}$

$x = \dfrac{16\sqrt{2}}{8} \:or\: \dfrac{-6\sqrt{2}}{8}$

$x = 2\sqrt{2} \: or \: \dfrac{-3\sqrt{2}}{4}$