Question

find the zeros of the polynomial of x^2-5x+3​

Answer 1

$\huge\underline{ \underline{ \bf{ \red{ \: solution \: : = }}}}$

$\bf {\blue{ {x }^{2} - 5x + 3 = 0 }}$

By using completing the square method for solving this equation.

move constant to the right hand side and change it's sign.

⠀⠀⠀⠀⠀$\bf {\green{ {x}^{2} - 5x = - 3 }}$

Add $\bf {\green{ (\frac{5}{2} ) {}^{2} }}$to both the sides of the equation.

$\bf {\green{ {x}^{2} - 5x + (\frac{5}{2} ) {}^{2} = - 3 + (\frac{5}{2} ) {}^{2} </strong><strong>}}</strong><strong>\</strong><strong>\</strong><strong>$

Using ,

⠀⠀⠀$\bf{ \boxed{ \purple{ \fbox{ {a}^{2} + {b}^{2} - 2ab = (a - b) {}^{2} }}}}$

⠀$\bf{ \green{(x - \frac{5}{2} ) {}^{2} = - 3 + \frac{5}{2} {}^{2} }}$

⠀$\bf {\green{(x - \frac{5}{2} ){}^{2} = \frac{13}{4} }}$

⠀⠀⠀⠀⠀$\bf{ \boxed{ \fbox {\purple{x = \frac{ - \sqrt{13} + 5}{2} }}}}$

⠀⠀⠀⠀⠀$\bf{ \boxed{ \fbox {\purple{x = \frac{ \sqrt{13} + 5 }{2} }}}}$

Answer 2

Note: I am denoting square root like this /.

zeroes of the polynomial are

-b+/(b^2-4ac)÷2a. (or)-b-/(b^2-4ac)÷2a

x^2-5x+3

(-(-5)+/(25-4(1)(3)))÷2(1) (or)(-(-5)-/(25-4(1)(3)))÷2(1)

(5+/(25-12))÷2 (or)(5-/(25-12))÷2

(5+/13)÷2 (or) (5-/13)÷2

Hope it helps you.