Question

If the zeroes of the polynomial ax2+bx+b=0 are in the ratio m:n, then find the value of root m/ root n + root n/root m.

Answer 1

See the attachment for detail solution

Answer 2

Answer:

$\frac{-\sqrt{b} }{\sqrt{a} }$ is the required value for  $\frac{\sqrt{m} }{\sqrt{n} } + \frac{\sqrt{n} }{\sqrt{m} }$.

Step-by-step explanation:

Explanation:

Given , $ax^{2}+ bx+b = 0$

zeroes are in ratio = m:n

Let , $\alpha$ and $\beta$ be the roots of the given polynomial .

Therefore , $\alpha :\beta = m:n$

Now, $\frac{\alpha }{\beta } = \frac{m}{n}$

⇒α= mk and $\beta$ = nk

Step 1:

Now from the given polynomial we have ,

$\alpha +\beta = \frac{-b}{a}$ and $\alpha \beta = \frac{b}{a}$

⇒m + n = $\frac{-b}{ak}$ and mn = $\frac{b}{ak^{2}}$

We have,  $\frac{\sqrt{m} }{\sqrt{n} } + \frac{\sqrt{n} }{\sqrt{m} }$  = $\frac{m+ n}{\sqrt{mn} }$

Now put the value of (m+n )and (mn )in the above equation,

⇒ $\frac{m+ n}{\sqrt{mn} }$ = $\frac{\frac{-b}{ak} }\sqrt{{\frac{b}{ak^{2} } }}$ =  $\frac{-b}{ak} .\frac{k\sqrt{a} }{\sqrt{b} }$  

(here , k and k is cancel out  )

∴$\frac{\sqrt{m} }{\sqrt{n} } + \frac{\sqrt{n} }{\sqrt{m} }$ = $\frac{-\sqrt{b} }{\sqrt{a} }$ .

Final answer :

Hence , the value of $\frac{\sqrt{m} }{\sqrt{n} } + \frac{\sqrt{n} }{\sqrt{m} }$ is $\frac{-\sqrt{b} }{\sqrt{a} }$ .

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