Question

I have 2 questions 1 in image and 1 is : A rational number which has non-terminating decimal representation is :- 91/32 , 101/625 , 49/105​

Answer 1

Answer:

ans -11/10 is correct answer

Answer 2

Answer:

1) -11/ 10

2) 49/ 105

Step-by-step explanation:

We'll have to solve for $\alpha$ and $\beta$ to find out the value of the required expression.

It says $\alpha$ and $\beta$ are the roots of the expression x² - 3x + 10, i.e., the values of x for which the value of the expression is 0.

==> x² - 3x + 10 = 0 . . . . . eqn. (¡)

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Nature of roots:

Let's first check the nature of roots, by using the discriminant properties of quadratic equations:

$\boxed{ \mathfrak{D(discrminant) = {b}^{2} - 4ac}}$

Where a, and b are the coefficients of x²and x respectively and c is the constant term in a quadratic equation.

for eqn. (¡):

• a = 1
• b = -3
• c = 10

D = b² - 4ac

= (-3)² - 4(1)(10)

= 9 - 40

= -31

D = -31

==> D < 0

If D < 0, the roots are imaginary numbers! number that can't be obtained on the Cartesian plane.

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Roots of the equation:

The roots of a quadratic equation are given by:

$\boxed{ \mathsf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

$\implies \mathsf{ x = \frac{ - ( - 3) \pm \sqrt{ {( - 3)}^{2} - 40 } }{2 \times 1} }$

$\implies \mathsf{ x = \frac{ 3 \pm \sqrt {9 - 40 }}{2} }$

√(-31) can be written as √(31) × √(-1)

and √(-1) = i

==> √(-31) = √(31) i

$\implies \mathsf{ x = \frac{ 3 \pm \sqrt { - 31}}{2} }$

$\implies \mathsf{ x = \frac{ 3 \pm \sqrt { 31} \: i}{2} }$

Therefore, the values of $\alpha$ and $\beta$ (roots of the equation) are:

$\: \mathsf{ \alpha = \frac{ 3 + \sqrt { 31} \: i}{2} }$

$\: \mathsf{ \beta = \frac{ 3 - \sqrt { 31} \: i}{2} }$

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Final step to the answer:

We, now, have to find out the value of

$\boxed{( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } )}$

Substituting the values of $\alpha$ and $\beta$ :

$\implies \mathsf{ (\frac{ \frac{3 + \sqrt{31 \: i} }{2} }{ \frac{3 - \sqrt{31} \: i }{2} } \: + \: \frac{ \frac{3 - \sqrt{31 \: i} }{2} }{ \frac{3 + \sqrt{31} \: i }{2} } )}$

2 gets canceled out:

$\implies \mathsf{ (\frac{ 3 + \sqrt{31 \: i} }{ {3 - \sqrt{31} \: i } } \: + \: \frac{ 3 - \sqrt{31 \: i } }{3 + \sqrt{31} \: i } )}$

taking LCM, and getting a common denominator:

$\implies \mathsf{ (\frac{3 + \sqrt{31}\: i)^{2} +( 3 - \sqrt{31} \: i ) ^{2} }{(3 - \sqrt{31} \: i)(3 + \sqrt{31} \: i) }}$

• (3 + √31 i)² + (3 - √31 i)² = 9 - 31 + 6√31 i+ 9 - 31 - 6√31 i

= 18 - 62

= - 44

[i² = -1]

[(a + b)² = a² + b² + 2ab]

• (3 + √31 i) (3 - √31 i) = 3² - 31 i²

(3 + √31 i) (3 - √31 i) = 3² - 31 i²= 9 + 31

(3 + √31 i) (3 - √31 i) = 3² - 31 i²= 9 + 31 = 40

[(a + b)(a - b) = a² - b²]

$\implies \mathsf{ \frac{ - 44 }{40 }}$

4 being common in both Numerator and denominator gets canceled out

= -11/ 10

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Answer:

$= \mathsf {\frac{ - 11}{10} }$

That is option 1!

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Second question, answer:

$\boxed{ \frac{49}{105} }$

which yields a value of 0.466..

while,

101/ 625 = 0.1616

91/ 32 = 2.84375

$\mathfrak{ \overline{there \: you \: are!}}$

Your questions are ingenious! Thank you! :)