Question

find the zeros of 4 x square minus 7 and verify the relationship between the zeros and its coefficient

Answer 1

Hello ,
$4 {x}^{2} - 7 \\ {(2x)}^{2} - {( \sqrt{7}) }^{2} \\ using \: {x}^{2} + {y}^{2} = (x + y)(x - y) \\ (2x + \sqrt{7} )(2x - \sqrt{7} ) \\ so \: \\ the \: zerores \: of \: the \: polynomial \: are \\ \frac{ - \sqrt{7} }{2} and \frac{ \sqrt{7} }{2}$

Now,
The polynomial is
$4 {x}^{2} - 7$
So,
Sum of zeroes = - b/a = 0/4 = 0
Also,
$- \frac{ \sqrt{7} }{2} + \frac{ \sqrt{7} }{2} = 0$
So, sum of zeroes = - b/a

Now,

Product of zeroes = d/a = - 7/4
Also,
$\frac{ \sqrt{7} }{2} \times ( - \frac{ \sqrt{7} }{2} ) = - \frac{7}{4}$
Hence
Product of zeroes = d/a

So,
The relation between the zeroes and the coefficients is verified.

Hope this will be helping you...

Answer 2

Step-by-step explanation:

होप दिस आंसर विल हेल्प योर इक्विलाइजेशन सर प्लीज लाइक दिस