Question

x^2-2√3-1=0
first dolve​

Answer 1

Step-by-step explanation:

Solution:−

Let,

p(x) = 3 {x}^{4} + 6 {x}^{3} - 2 {x}^{2} - 10x - 5p(x)=3x

4

+6x

3

−2x

2

−10x−5

Given,

\sqrt{ \frac{5}{3} } \: \: and \: \: - \sqrt{ \frac{5}{3} }

3

5

and−

3

5

are roots of p( x )

So,

(x - \sqrt{{ \frac{5}{3} } } )(x−

3

5

)

(x + \sqrt{ \frac{5}{3} })(x+

3

5

) are factors of p( x )

Also,

(x + \sqrt{ \frac{5}{3} } )(x - \sqrt{ \frac{5}{3} } )(x+

3

5

)(x−

3

5

) is a factor of p( x )

i.e

{x}^{2} - \frac{5}{3}x

2

−

3

5

is a factor of p ( x )

Divide p( x ) with it's factor

\frac{3 {x}^{4} + 6 {x}^{3} - 2 {x}^{2} - 10x - 5 }{ {x}^{2} - \frac{5}{3} }

x

2

−

3

5

3x

4

+6x

3

−2x

2

−10x−5

See the attachment

It gives

3 {x}^{2} + 6x + 33x

2

+6x+3

Factorize it

3 \times 3 = 93×3=9

3 + 3 = 63+3=6

So,

3 {x}^{2} + 3x + 3x + 33x

2

+3x+3x+3

3x(x + 1) + 3(x + 1)3x(x+1)+3(x+1)

(x + 1)(3x +3 )(x+1)(3x+3)

x + 1 = 0x+1=0

x = - 1x=−1

And

3x + 3 = 03x+3=0

3x = - 33x=−3

x = - \frac{ 3}{3} = - 1x=−

3

3

=−1

The other factors are -1 and -1