Question

Solve the attached MCQ question please, I'll mark the Brainliest when I return :)​

Answer 1

Question 8)

Answer:

Given: (1 + x + x²) / (1 - x + x²) = 62(1 + x) / 63(1 - x)

To find: x

Solution:

(1 + x + x²) / (1 - x + x²) = 62(1 +x) / 63(1 - x)

(Cross multiplying: )

=> 63(1 - x)(1 + x +x²) = 62(1 + x)(1 - x + x²)

=> 63[1 + x+ x² - x - x² - x³] = 62 [ 1 - x + x² + x- x² + x³]

=> 63[1 - x³] = 62[1 + x³]

=> 63 - 63x³ = 62 + 62x³

=> 63 - 62 = 62x³ + 63x³

=> 1 = 125x³

x³ = 1/125

x³ = (1/5)³

∴ x = 1/5

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Question 9)

If (x + 2) and (x - 3) are factors of x² + ax + b, find a and b.

Answer:

p(x) = x² + ax + b

Two of the factors are (x + 2) and (x - 3).

∴ Let x be -2 and 3 respectively.

Put the values of x in given equation:

(x = -2)

p(x) = (-2)² + a(-2) + b

0 = 4 - 2a + b

-4 = - 2a + b .... (1)

(x = 3)

p(x) = (3)² + a(3) + b

0 = 9 + 3a + b

-9 = 3a + b .... (2)

On subtracting (1) and (2):

-4 = -2a + b

- -9 = 3a + b

____________

5 = -5a

-1 = a

Putting value of a in equation (1),

- 4 = - 2(-1) + b

- 4 - 2 = b

- 6 = b

Thus, value of a is -1, and that of b is -6.

Answer 2

Step-by-step explanation:

$\huge{\underline{\red{AnswEr}}}$

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8)

$\implies \frac{1 + x + {x}^{2} }{1 - x + {x}^{2} } = \frac{62(1 + x)}{63(1 - x)}$

$\implies63(1 - x)(1 + x + {x}^{2} ) = 62(1 + x)(1 - x + {x}^{2} )$

$\implies63(1 + x + {x}^{2} - x - {x}^{2} - {x}^{3} ) = 62(1 - x + {x}^{2} + x - {x}^{2} + {x}^{3} )$

$\implies63(1 - {x}^{3} ) = 62(1 + {x}^{3} )$

$\implies63 - {63x}^{3} = 62 + 62x ^{3}$

$\implies63 - 62 = 63 {x}^{3} + 62 {x}^{3}$

$\implies1 = 125 {x}^{3}$

$\implies {x}^{3} = \frac{1}{125}$

$\implies \: x = \sqrt[3]{ \frac{1}{125} }$

$\implies \: x = \frac{1}{5}$