Question

Find the zeros of quadratic polynomial:
$4x^{2} -6 -8x$

Answer 1

Answer:

Step-by-step explanation:

The zeroes of given polynomial are

$x_{1}$ = 1 - $\frac{\sqrt{10} }{2}$

$x_{2}$ = 1 + $\frac{\sqrt{10} }{2}$

Answer 2

Step-by-step explanation:

●Method l :

Let p(x) = 4x² - 6 - 8x

Now,

4x² - 6 - 8x

= 4x² - 8x - 6

= (2x)² - 2.2x.4/2 + (4/2)² - (4/2)² - 6

= (2x - 4/2)² - (4/2)² - 6

= ( 2x - 2 )² - 2² - 6

= ( 2x - 2 )² - 4 - 6

= ( 2x - 2 )² - 10

To find the zeroes of p(x), we write p(x) = 0

$\bold{ \tt{ {(2x - 2)}^{2} - 10= 0}} \\ \\ \bold{ \tt{ = > \: \: \: \: \: \: \: \: \: \: (2x - 2)^{2} = 10}} \: \: \: \: \: \\ \\ \bold{ \tt{ \: \: \: \: \: = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x - 2= { \pm}\sqrt{10} }} \: \: \: \\ \\ \bold {\tt{ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x = 2\pm\sqrt{10} }} \\ \\ \bold {\tt{ \: = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ 2 \pm\sqrt{10} }{2} }} \\ \\ \bold{ \tt{ \: \: \: \: \: \: \: \: \therefore \: x \: = \frac{ 2 + \sqrt{10} }{2} , \: \frac{ 2- \sqrt{10} }{2} }}$

Therefore, the zeroes of the quadratic polynomial 4x² - 6 - 8x are ( 2 + √10 )/2 and ( 2 - √10 )/2 .

●Method ll :

Comparing 4x² - 6 - 8x to ax² + bx + c , we get

a = 4

b = -8

c = -6

D = b² - 4 ac

= (-8)² - 4 × 4 × (-6)

= 64 + 96

= 160

Using quadratic formula,

$\bold{\tt{x = \frac{ - b \pm\sqrt{D} }{2a} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\tt{ = \frac{ - ( - 8) \pm \sqrt{160} }{2 \times 4}}} \\ \\ \bold{\tt{ = \frac{ 8 \pm4 \sqrt{10} }{2 \times 4}}} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\tt{ = \frac{ \cancel4( 2 \pm \sqrt{10}) }{2 \times \cancel 4} }} \: \: \: \: \: \: \\ \\ \bold{\tt{ \: \: \: \: \: \: \: \: = \frac{ 2 + \sqrt{10} }{2}, \frac{2 - \sqrt{10} }{2} }}$

Therefore, the zeroes of the quadratic polynomial are ( 2 + √10 )/2 and ( 2 - √10 )/2 .