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

Answer:

=>root b/a

Step-by-step explanation:

let the zeros of the given polynomial ax^2+bx+c be m"alpha" and n"alpha"

m "alpha " + n"alpha" = -b/a

=>"alpha"= -b/a(m+n)------------------(i)

and similarly product of zeros  

=> mn"alpha^2"=b/a-------------(ii)

Now,

Putting the value of equation (i) in equation (ii)

You would get the desired answer which is  

root m/ root n +root n/root m

=>root b/a

Answer 2

Step-by-step explanation:

Given:

If the zeroes of the polynomial ax²+bx+b=0 are in the ratio m:n.

To find: find the value of

$\frac{ \sqrt{m} }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{m} }$

Solution:

Let

$\alpha \: \text{and} \: \beta$

are the roots of given quadratic equation.

Write the relation between coefficient of equation and zeroes of equation.

$\alpha + \beta = \frac{ - b}{a} \: ...eq1 \\ \\ \alpha \beta = \frac{c}{a} \: ...eq2$

ATQ

Zeroes of polynomial are in the ratio of m:n,that means one factor is common in both.Let p is the common factor,so

$\alpha = pm \\ \beta = pn$

put these values into eq1 and eq2

$\alpha + \beta = p(m + n) = \frac{ - b}{a} \\ \\ m + n = \frac{ - b}{ap} \: ...eq3$

$\alpha \beta = {p}^{2} mn = \frac{b}{a} \\ \\ mn = \frac{b}{a {p}^{2} } \: ...eq4$

To find the value of

$\frac{ \sqrt{m} }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{m} }$

simplify it,by taking LCM and simplify

$\frac{ \sqrt{m} }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{m} } = \frac{ \sqrt{m} \sqrt{m} + \sqrt{n} \sqrt{n} }{ \sqrt{m} \sqrt{n} } \\ \\ \implies \frac{m + n}{ \sqrt{mn} }$

Put value from eq 3 and eq4

$\frac{ m + n}{ \sqrt{mn} } = \frac{ \frac{ - b}{ap} }{ \sqrt{ \frac{b}{a {p}^{2} } } } \\ \\ \implies \frac{ - b \: \times p \: \sqrt{a} }{ap \times \sqrt{b} } \\ \\ \implies \frac{ - b}{ \sqrt{a} \sqrt{b} } \\ \\ \implies \frac{ - \sqrt{b}}{ \sqrt{a}}$

Thus,

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

Final answer:

$\boxed{ \bold{\frac{ \sqrt{m} }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{m} } = -\sqrt{\frac{ b}{a}} }}$

Hope it helps you.

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