Question

if p and q are the roots of the equation 2x^2-5x+8 then the value of p/q+q/p​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\:p \: and \: q \: are \: roots \: of \: {2x}^{2} - 5x + 8 = 0}$

So, We know that

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:pq = \dfrac{8}{2} = 4$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:p + q = - \dfrac{( - 5)}{2} = \dfrac{5}{2}$

Now, Consider

$\rm :\longmapsto\: {p}^{2} + {q}^{2}$

$\rm \:  =  \: {p}^{2} + {q}^{2} + 2pq - 2pq$

$\rm \:  =  \: ({p}^{2} + {q}^{2} + 2pq) - 2pq$

$\rm \:  =  \: ({p + q)}^{2} - 2pq$

$\rm \:  =  \: {\bigg[\dfrac{5}{2} \bigg]}^{2} - 2 \times 4$

$\rm \:  =  \:\dfrac{25}{4} - 8$

$\rm \:  =  \:\dfrac{25 - 32}{4}$

$\rm \:  =  \: - \: \dfrac{7}{4}$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \: {p}^{2} + {q}^{2} = \: - \: \frac{7}{4} \: \: }}}$

So,

$\purple{\rm :\longmapsto\:\dfrac{p}{q} + \dfrac{q}{p} }$

$\purple{\rm \:  =  \:\dfrac{ {p}^{2} + {q}^{2} }{pq} \: }$

On substituting the values, evaluated above, we get

$\purple{\rm \:  =  \: - \: \dfrac{7}{4} \times \dfrac{1}{4} \: }$

$\purple{\rm \:  =  \: - \: \dfrac{7}{16} \: }$

Thus :-

$\purple{\rm :\longmapsto\:\boxed{ \tt{ \: \: \: \dfrac{p}{q} + \dfrac{q}{p} } = - \: \frac{7}{16} \: \: \: }}$

Additional Information :-

More Identities to know :-

$\boxed{ \tt{ \: {p}^{2} + {q}^{2} = {(p + q)}^{2} - 2pq \: }}$

$\boxed{ \tt{ \: {p}^{3} + {q}^{3} = {(p + q)}^{3} - 3pq(p + q) \: }}$

$\boxed{ \tt{ \: {(p - q)}^{2} = {(p + q)}^{2} - 4pq \: \: }}$

$\boxed{ \tt{ \: {p}^{4} + {q}^{4} = {\bigg( {(p + q)}^{2} - 2pq\bigg) }^{2} - 2 {(pq)}^{2} }}$

Answer 2

Step-by-step explanation:

Solution−

Given that,

\red{\rm :\longmapsto\:p \: and \: q \: are \: roots \: of \: {2x}^{2} - 5x + 8 = 0}:⟼pandqarerootsof2x

2

−5x+8=0

So, We know that

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

Product of the zeroes=

coefficient of x

2

Constant

\bf\implies \:pq = \dfrac{8}{2} = 4⟹pq=

2

8

=4

And

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

Sum of the zeroes=

coefficient of x

2

−coefficient of x

\bf\implies \:p + q = - \dfrac{( - 5)}{2} = \dfrac{5}{2}⟹p+q=−

2

(−5)

=

2

5

Now, Consider

\rm :\longmapsto\: {p}^{2} + {q}^{2}:⟼p

2

+q

2

\rm \: = \: {p}^{2} + {q}^{2} + 2pq - 2pq = p

2

+q

2

+2pq−2pq

\rm \: = \: ({p}^{2} + {q}^{2} + 2pq) - 2pq = (p

2

+q

2

+2pq)−2pq

\rm \: = \: ({p + q)}^{2} - 2pq = (p+q)

2

−2pq

\rm \: = \: {\bigg[\dfrac{5}{2} \bigg]}^{2} - 2 \times 4 = [

2

5

]

2

−2×4

\rm \: = \:\dfrac{25}{4} - 8 =

4

25

−8

\rm \: = \:\dfrac{25 - 32}{4} =

4

25−32

\rm \: = \: - \: \dfrac{7}{4} = −

4

7

Hence,

\red{\rm \implies\:\boxed{ \tt{ \: \: {p}^{2} + {q}^{2} = \: - \: \frac{7}{4} \: \: }}}⟹

p

2

+q

2

=−

4

7

So,

\purple{\rm :\longmapsto\:\dfrac{p}{q} + \dfrac{q}{p} }:⟼

q

p

+

p

q

\purple{\rm \: = \:\dfrac{ {p}^{2} + {q}^{2} }{pq} \: } =

pq

p

2

+q

2

On substituting the values, evaluated above, we get

\purple{\rm \: = \: - \: \dfrac{7}{4} \times \dfrac{1}{4} \: } = −

4

7

×

4

1

\purple{\rm \: = \: - \: \dfrac{7}{16} \: } = −

16

7

Thus :-

\purple{\rm :\longmapsto\:\boxed{ \tt{ \: \: \: \dfrac{p}{q} + \dfrac{q}{p} } = - \: \frac{7}{16} \: \: \: }}:⟼

q

p

+

p

q

=−

16

7

Additional Information :-

More Identities to know :-

\boxed{ \tt{ \: {p}^{2} + {q}^{2} = {(p + q)}^{2} - 2pq \: }}

p

2

+q

2

=(p+q)

2

−2pq

\boxed{ \tt{ \: {p}^{3} + {q}^{3} = {(p + q)}^{3} - 3pq(p + q) \: }}

p

3

+q

3

=(p+q)

3

−3pq(p+q)

\boxed{ \tt{ \: {(p - q)}^{2} = {(p + q)}^{2} - 4pq \: \: }}

(p−q)

2

=(p+q)

2

−4pq

\boxed{ \tt{ \: {p}^{4} + {q}^{4} = {\bigg( {(p + q)}^{2} - 2pq\bigg) }^{2} - 2 {(pq)}^{2} }}

p

4

+q

4

=((p+q)

2

−2pq)

2

−2(pq)

2