Question

Q. If k is the ratio of the roots of the equation x² - ax + b=0, then the value of $\frac{ {k}^{2} + 1}{k}$

Answer 1

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♦♦ Quadratic Resolution ♦♦

→ ( x² - ax + b ) = 0

→ Let the roots be : x₁ and x₂ , and 'k' be their ratio

◙ Sum of Roots = -( coefficient of x ) ÷ ( coefficient of x² ) = -( - a ) = a

◙ Product of Roots = ( constant term ) ÷ ( coefficient of x² ) = b

=> ( Sum of Roots )² ÷ ( Product of Roots ) = ( a² ÷ b )

$-\ \textgreater \ \frac{( x_1 + x_2 )^2 }{x_1 x_2} = \frac{a^2}{b} \\ \\ =\ \textgreater \ \frac{ x_1^2 + x_2^2 + 2x_1 x_2 }{x_1 x_2} = \frac{a^2}{b} \\ \\ =\ \textgreater \ \frac{x_1}{x_2} + \frac{x_2}{x_1}+ 2 = \frac{a^2}{b} \\ \\ =\ \textgreater \ k + \frac{1}{k} = \frac{a^2 - 2b}{b}$

=> ( k² + 1 ) ÷ k  = ( a² - 2b ) / b
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→ Just Ping me anytime :v:

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that,

→ x² - ax + b = 0

→ k  is the ratio of their roots.

To find,

The value of $\frac{k^{2}+1}{k}$

Solution:

Let the roots of the equation be α , β respectively.

And general form of a quadratic equation is px² + qx + r = 0

In the given equation p = 1, q = (-a) , r = b

Then , We know that,

$Sum\ of\ roots = \frac{-q}{p}$

⇒ $\alpha + \beta = \frac{-(-a)}{1} = \frac{a}{1} = a$

$Product\ of\ roots = \frac{r}{p}$

⇒ $\alpha \times \beta = \frac{b}{1} = b$

$Ratio\ of\ roots = \frac{ \alpha }{ \beta } = k$

Now, 

$\frac{k^{2} + 1}{k} = k + \frac{1}{k}$

Now substitute the value of k, 

$k + \frac{1}{k} = \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

⇒ $\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{( \alpha )^{2}+( \beta )^{2} }{( \alpha \beta )}$

⇒  $\frac{( \alpha)^{2} + ( \beta )^{2}}{( \alpha \beta )} = \frac{( \alpha + \beta )^{2} - 2( \alpha \beta )}{ \alpha \beta }$

Now substitute the values of ( α + β) & (αβ)

 ⇒ $\frac{( \alpha + \beta )^{2} - 2( \alpha \beta )}{ \alpha \beta } = \frac{(a)^{2}-2(b)}{b}$

Therefore: $\bold{\frac{k^{2}+1}{k} = \frac{a^{2}-2b}{b} }$
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Hope my answer is helpful to you.