Question

please answer for the questions in attatchement
and get yourself marked as brainliest
no irrelavent answers please..
from 16 to 30.
​

Answer 1

Answer:please mark me as brainliest

Step-by-step explanation:

16)Answer:

Solution :-

Here, we have

9x² - 9(a + b)x + (2a² + 5ab + 2b²) = 0

Therefore,

Constant term = 2a² + 5ab + 2b²

= 2a² + 4ab + ab + 2b²

= 2a(a + 2b) + b(a + 2b)

= (a + 2b) (2a + b)

Coefficient of middle term = - 9(a + b) = - 3[(2a + b) + (a + 2b)

Then,

⇒ 9x² - 9(a + b)x + (2a² + 5ab + 2b²) = 0

⇒ 9x² - 3[(2a + b) + (a + 2b)]x + (2a + b) (a + 2b) = 0

⇒ 9x² - 3(2a + b)x - 3(a + 2b)x + (2a + b) (a + 2b) = 0

⇒ 3x[3x - (2a + b) - (a + 2b) [3x - (2a + b)] = 0

⇒ [3x - (2a + b)] [3x - (a + 2b) = 0

⇒ [3x - (2a +b)] = 0 or [3x - (a + 2b) = 0

⇒ x = 2a + b/3, a + 2b/3

Hence, x = 2a + b/3, a + 2b

17) in attachment

18) sorry i am not able to find

19)in attachment

16) ii)a²b²x² - (4b⁴ - 3a⁴)x - 12a²b² = 0

Here A=a²b²

B= -(4b⁴ - 3a⁴)

C= -  12a²b²

D=discriminant =B²-4AC=

=[( -(4b⁴ - 3a⁴)² - 4 x [a²b²] x -  12a²b²]

=[16b⁸ + 9a⁸ -24a⁴b⁴]+ 48 a⁴ b⁴

=16b⁸ + 9a⁸+24 a⁴ b⁴

=(4b⁴)² +2x 4b x 3a ² + (3a²)²

=(4b⁴+3a⁴)²     [∵ (a+b)²=a²+b²+2ab]

Now, let us use quadratic formula 

X = {-B ± √(B²-4AC)}/2A

= -( -(4b⁴ - 3a⁴)  ±√ (4b⁴+3a⁴)² /2a²b²

={(4b⁴ - 3a⁴)  ± (4b⁴+3a⁴)} /2a²b²

now taking positive sign,

=[(4b⁴ - 3a⁴) + 4b⁴ +3a⁴]/2a²b²

=8b⁴/2a²b²

x=4b²/a²

taking negative value :

=[(4b⁴ - 3a⁴) - 4b⁴ -3a⁴]/2a²b²

=-6a⁴/2a²b²

= -3a²/b²

∴X=4b²/a² ,  -3a²/b^2

20)f you want to do it by trial and error, which is how I did it the answer is

(2x - 3)(3x + 5)= 0

Hence, x = 3/2, -5/3

A more systematic approach:

Multiply 6 and -15: 6*-15 = -90

Look for 2 integers that multiply to give -90 and add to give 1: 10, -9

Hence, 6x^2 + x - 15

= 6x^2 + 10x - 9x - 15

= 3x(2x + 5) - 3(2x + 5)

=(2x - 3)(3x + 5) = 0

Hence, x = 3/2, -5/3

21)The first term is,  12x2  its coefficient is  12 .

The middle term is,  -x  its coefficient is  -1 .

The last term, "the constant", is  -1  

Step-1 : Multiply the coefficient of the first term by the constant   12 • -1 = -12  

Step-2 : Find two factors of  -12  whose sum equals the coefficient of the middle term, which is   -1 .

     -12    +    1    =    -11  

     -6    +    2    =    -4  

     -4    +    3    =    -1    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  3  

                    12x2 - 4x + 3x - 1

Step-4 : Add up the first 2 terms, pulling out like factors :

                   4x • (3x-1)

             Add up the last 2 terms, pulling out common factors :

                    1 • (3x-1)

Step-5 : Add up the four terms of step 4 :

                   (4x+1)  •  (3x-1)

            Which is the desired factorization

22)in attachment

23)sorry i am not able to solve the question

24)4√3 x² +5x -2√3 =0

⇒4√3 x² +8x -3x -2√3=0

⇒4x(√3 x+2) -√3(√3x +2) =0

⇒(√3x +2)(4x -√3) =0

⇒(√3x +2) =0 or (4x-√3) =0

⇒x = -2/√3 or √3/4

25) in attachments

26)Let --> y = ( a / ( a + b ) ) ;

--> x² + [ y + 1/y ]x + 1 = 0

=> yx² + [ y² + 1 ] x + y = 0

=> yx² + y²x + x + y = 0

=> [ yx + 1 ][ x + y ] = 0

=> x = [ - 1 / y ]  || x = - y

=> x = [ ( - a - b ) / a ] || x = [ -a / ( a + b )

27)4x²-4a²x+(a⁴-b⁴)=0

or, 4x²-2{(a²+b²)+(a²-b²)}x+(a²+b²)(a²-b²)=0

or, 4x²-2(a²+b²)x-2(a²-b²)x+(a²+b²)(a²-b²)=0

or, 2x{2x-(a²+b²)}-(a²-b²){2x-(a²+b²)}=0

or, {2x-(a²+b²)}{2x-(a²-b²)}=0

Either, 2x-(a²+b²)=0

or, 2x=a²+b²

or, x=(a²+b²)/2

Or, 2x-(a²-b²)=0

or, 2x=a²-b²

or, x=(a²-b²)/2

∴, x=(a²+b²)/2, (a²-b²)/2 Ans.

28)in attachment

29,30,23 ,18 sorry i am not able to answer