Math, asked by Manigagan, 10 months ago

If 2x=3+√5 and y=x^3 then y satisfies the quadratic equation (i) y^2 -18y+1​

Answers

Answered by lublana
1

Answer with Step-by-step explanation:

2x=3+\sqrt 5

x=\frac{3+\sqrt 5}{2}

y=x^3

Quadratic equation :y^2-18y+1

 Substitute the values then we get

(x^3)^2-18x^3+1

x^6-18x^3+1

x^3(x^3-18)+1

(\frac{3}{2}+\frac{\sqrt 5}{2})^3((\frac{3}{2}+\frac{\sqrt 5}{2})^3-18)+1

Using identity:(a+b)^3=a^3+b^3+3a^2b+3ab^2

(\frac{27}{8}+\frac{5\sqrt 5}{8}+\frac{27\sqrt 5}{8}+\frac{45}{8})((\frac{27}{8}+\frac{5\sqrt 5}{8}+\frac{27\sqrt 5}{8}+\frac{45}{8}-18)+1

(\frac{27+45}{8}+\frac{27\sqrt 5+5\sqrt 5}{8})(\frac{27+45}{8}+\frac{27\sqrt 5+5\sqrt 5}{8}-18)+1

(9+\frac{32\sqrt 5}{8})(9+\frac{32\sqrt 5}{8}-18)+1

(9+\frac{32\sqrt 5}{8})(\frac{32\sqrt 5}{8}-9)+1

(\frac{32\sqrt 5}{8})^2-9^2+1

Using identity:(a+b)(a-b)=a^2-b^2

80-81+1=0

LHS=RHS

Hence,y satisfies the quadratic equation.

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https://brainly.in/question/12770535:Answered by Spiderman                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                      

Answered by shobaranikaserevene
0

Answer:

x=

2

3+

5

y=x^3y=x

3

Quadratic equation :y^2-18y+1y

2

−18y+1

Substitute the values then we get

(x^3)^2-18x^3+1(x

3

)

2

−18x

3

+1

x^6-18x^3+1x

6

−18x

3

+1

x^3(x^3-18)+1x

3

(x

3

−18)+1

(\frac{3}{2}+\frac{\sqrt 5}{2})^3((\frac{3}{2}+\frac{\sqrt 5}{2})^3-18)+1(

2

3

+

2

5

)

3

((

2

3

+

2

5

)

3

−18)+1

Using identity:(a+b)^3=a^3+b^3+3a^2b+3ab^2(a+b)

3

=a

3

+b

3

+3a

2

b+3ab

2

(\frac{27}{8}+\frac{5\sqrt 5}{8}+\frac{27\sqrt 5}{8}+\frac{45}{8})((\frac{27}{8}+\frac{5\sqrt 5}{8}+\frac{27\sqrt 5}{8}+\frac{45}{8}-18)+1(

8

27

+

8

5

5

+

8

27

5

+

8

45

)((

8

27

+

8

5

5

+

8

27

5

+

8

45

−18)+1

(\frac{27+45}{8}+\frac{27\sqrt 5+5\sqrt 5}{8})(\frac{27+45}{8}+\frac{27\sqrt 5+5\sqrt 5}{8}-18)+1(

8

27+45

+

8

27

5

+5

5

)(

8

27+45

+

8

27

5

+5

5

−18)+1

(9+\frac{32\sqrt 5}{8})(9+\frac{32\sqrt 5}{8}-18)+1(9+

8

32

5

)(9+

8

32

5

x= 3+√5/2 & y=x³, then y satisfies the quadratic equation

−18)+1

(9+\frac{32\sqrt 5}{8})(\frac{32\sqrt 5}{8}-9)+1(9+

8

32

5

)(

8

32

5

−9)+1

(\frac{32\sqrt 5}{8})^2-9^2+1(

8

32

5

)

2

−9

2

+1

Using identity:(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a

2

−b

2

80-81+1=080−81+1=0

LHS=RHS

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