Question

pls ans to be brainlist

Answer 1

Let the given equation be f(x) = 4x^2 - 8x - 6.

Now,

We need to find the zero of the polynomial.

Put f(x) = 0

4x^2 - 8x - 6 = 0

Compare with ax^2 + bx + c = 0, we get a = 4, b = -8, c = -6.

(1)

$x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \frac{-(-8) + \sqrt{(-8)^2 - 4 * 4 * (-6)} }{2(4)}$

$= \frac{8 + \sqrt{160} }{8}$

$= \frac{8 + 4 \sqrt{10} }{8}$

$= \frac{4(2 + \sqrt{10} )}{8}$

$= \frac{2 + \sqrt{10} }{2}$



(2)

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

$= \frac{-(-8) - \sqrt{(-8)^2 - 4 * 4 * (-6)} }{2 * 4}$

$= \frac{8 - \sqrt{160} }{8}$

$= \frac{8 - 4 \sqrt{10} }{8}$

$= \frac{4(2 - \sqrt{10} )}{8}$

$= \frac{2 - \sqrt{10} }{2}$


Therefore the zeroes of the given quadratic polynomial are : 

$\frac{2 + \sqrt{10} }{2} , \frac{2 - \sqrt{10} }{2}$


Verification:

(1)

Sum of roots : 

$\frac{2 + \sqrt{10} }{2} + \frac{2 - \sqrt{10} }{2}$

$\frac{2 + \sqrt{10} + 2 - \sqrt{10} }{2}$

$\frac{4}{2}$

2.


Sum of roots:

= -b/a

= -(-8)/4

= 8/4

= 2.



(2)

Product of roots: 

$\frac{2 + \sqrt{10} }{2} * \frac{2 - \sqrt{10} }{2}$

$\frac{(2 + \sqrt{10} )(2 - \sqrt{10}) }{2 * 2}$

$\frac{2^2 - ( \sqrt{10})^2 }{4}$

$\frac{4 - 10}{4}$

$\frac{-6}{4}$

$\frac{-3}{2}$


Product of roots:

= c/a

= -6/4

= -3/2.


Hence the relationship between zeroes and coefficients is verified.


Hope this helps!