Question

please answer all of them ​

Answer 1

Question:-

11. If 4x² + 9y² = 136 and xy = 10, find 2x + 3y.

Solution:-

We already have:-

• 4x² + 9y² = 136
• xy = 10 .... (a)

Also we know:-

• x² + y² = (x² - y²) + 2xy .... (i)

Now,

4x² + 9y² = 136

= (2x)² + (3y)² = 136

Applying the formula from (i) we get:-

= (2x - 3y)² + 12xy = 136

Expanding (2x - 3y) we get:-

= (2x)² + (3y)² - 12xy + 12xy = 136

= (2x)² + (3y)² + 12xy - 12xy = 136

= {(2x)² + (3y)² + 2 × 2x × 3y} - 12xy = 136

= (2x + 3y)² - 12xy = 136

Putting xy = 10 from (a)

= (2x + 3y)² - 12 × 10 = 136

= (2x + 3y)² - 120 = 136

= (2x + 3y)² = 136 + 120

= (2x + 3y)² = 256

= 2x + 3y = √256

= 2x + 3y = 16

∴ The value of 2x + 3y is 16.

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Question:-

12. If x + y + z = 9 and x² + y² + z² = 35, find the value of x³ + y³ + z³ - 3xyz.

Solution:-

We are already given:-

• x + y + z = 9 ------- (i)
• x² + y² + z² = 35 ------- (ii)

We also know:-

• (x + y + z)² = x² + y² + z² + 2(xy + yz + zx) .... (a)
• x³ + y³ + z³ - 3xyz = (x + y + z)[(x² + y² + z²) - (xy + yz + zx) .... (b)

Now,

From (i)

x + y + z = 9

Squaring on both sides:-

(x + y + z)² = (9)²

Using identity (a):-

= x² + y² + z² + 2(xy + yz + zx) = 81

= (x² + y² + z²) + 2(xy + yz + zx) = 81

From (ii):-

= 35 + 2(xy + yz + zx) = 81

= 2(xy + yz + zx) = 81 - 35

= 2(xy + yz + zx) = 46

=> xy + yz + zx = 46/2

=> xy + yz + zx = 23 ---------- (iii)

Now,

Putting all the values in identity (b):-

x³ + y³ + z³ - 3xyz = (x + y + z)[(x² + y² + z²) - (xy + yz + zx)]

= x³ + y³ + z³ - 3xyz = 9(35 - 23)

= x³ + y³ + z³ - 3xyz = 9 × 12

= x³ + y³ + z³ - 3xyz = 108

∴ The value of x³ + y³ + z³ - 3xyz is 108.

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Question:-

13. Factorise 16√5 x² - 50x + 5√5

Solution:-

By middle term factorisation:-

= 16√5x² - 40x - 10x + 5√5

= 8√5x(2x - √5) - 5(2x - √5)

= (2x - √5)(8√5x - 5)

Hence Factorised!

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Question:-

14. Factorise 27y³ + 125z³

Solution:-

= 27y³ + 125z³

= (3y)³ + (5z)³

Now using the identity:-

• a³ + b³ = (a - b)(a² + ab + b²)

= (3y + 5z)[(3y)² + 3y × 5z + (5z)²]

= (3y + 5z)(9y² + 15xy + 125z²)

Hence, Factorised!

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Question:-

15. Show that (x - 1) is a factor of the polynomial p(x) = x³ + 2x² - x - 2 and hence factorise p(x)

Solution:-

We have:-

• p(x) = x³ + 2x² - x - 2
• g(x) = (x - 1)

Now:-

g(x) = x - 1 = 0

= x = 1

Putting x = 1 in p(x)

p(1) = (1)³ + 2(1)² - 1 - 2

p(1) = 1 + 2 - 1 - 2

p(1) = 3 - 3

p(1) = 0

∴ (x - 1) is a factor of p(x) = x³ + 2x² - x - 2.

Now, factorising p(x)

= x³ + 2x² - x - 2

= x²(x + 2) - 1(x + 2)

= (x + 2)(x² - 1)

= (x+2)(x+1)(x-1)

Hence Factorised!

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