Question

the product of the zeros of the quadratic polynomial px²+5x+6 is 3 then the value of p is​

Answer 1

zeroes of a polynomial means f(x)=0

f(x)= px²+5x+6

therefore, px²+5x+6=0

by quadratic formula-

we can show that product of zeroes of quadratic polynomial is c/a

quadratic equation - ax²+bx+c

here c = 6, a= p

c/a = 3

or, 6/p=3

therefore, p=2

To show that product of zeroes of quadratic polynomial is c/a

$x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} or \: \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \: product \: = \\ (\frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a})(\frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}) \\ \\ = \frac{ - (- b + \sqrt{ {b}^{2} - 4ac })(b + \sqrt{ {b}^{2} - 4ac} }{4 {a}^{2} } \\ \\ \star{{a}^{2} - {b}^{2} = (a + b)(a - b)} \\ \\ = \frac{ - ( \sqrt{ {b}^{2} - 4ac } - {b})( \sqrt{ {b}^{2} - 4ac } + b) }{4 {a}^{2} } \\ \\ = \frac { - ( { \sqrt{ {b}^{2} - 4ac} }^{2} - {b}^{2}) }{4 {a}^{2} } \\ \\ = \frac{ - ( \cancel{{b}^{2}} - 4ac \cancel{- {b}^{2}}) }{4 {a}^{2} } \\ \\ = \frac{ \cancel4 \cancel{a}c}{ \cancel4 \cancel{{a}^{2}} \: a } \\ \\ = \frac{c}{a}$