If one root of 5x ^ 2 + 26x + k = 0 is reciprocal of the other, then what is the value of k?

If one zero of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other. So `beta = 1/alpha ⇒`

Now we have

`alphaxxbeta= (\text{Constant term})/(\text{Coefficient of}x^2)`

`= k /5`

Since`alphabeta =1`

Therefore we have

`alpha beta = k/5`

`1= k/5`

`⇒ k = 5`

Hence, the correct choice is `(b)`.

Answer

Verified

Hint: Given equation is a quadratic equation. According to the given condition if one root is reciprocal or other then their product will be 1. So first we will find the roots and then we will use this hint. Find the roots as \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:

Given equation is \[5{x^2} + 13x + k = 0\] of the form \[a{x^2} + bx + c = 0\].Thus a=5 ,b=13 and c=k.Now let’s find the roots \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]Putting the values,\[ \Rightarrow \dfrac{{ - 13 \pm \sqrt {{{\left( { - 13} \right)}^2} - 4 \times 5 \times k} }}{{2 \times 5}}\]\[ \Rightarrow \dfrac{{ - 13 \pm \sqrt {169 - 20k} }}{{10}}\]Now the two roots are \[ \Rightarrow \dfrac{{ - 13 + \sqrt {169 - 20k} }}{{10}}\] and \[\dfrac{{ - 13 - \sqrt {169 - 20k} }}{{10}}\]Given that the product of one of them and reciprocal of other. So,\[\dfrac{{ - 13 + \sqrt {169 - 20k} }}{{10}} = \dfrac{1}{{\dfrac{{ - 13 - \sqrt {169 - 20k} }}{{10}}}}\]Now we will solve above equation\[ \Rightarrow \dfrac{{ - 13 + \sqrt {169 - 20k} }}{{10}} = \dfrac{{10}}{{ - 13 - \sqrt {169 - 20k} }}\]Now cross multiply\[ \Rightarrow \dfrac{{ - 13 + \sqrt {169 - 20k} }}{{10}} \times \dfrac{{ - 13 - \sqrt {169 - 20k} }}{{10}} = 1\]In numerator we observe that it is \[\left( {a + b} \right)\left( {a - b} \right)\] so we can use the identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]\[ \Rightarrow \dfrac{{{{\left( { - 13} \right)}^2} - {{\left( {\sqrt {169 - 20k} } \right)}^2}}}{{100}} = 1\]\[ \Rightarrow \dfrac{{169 - \left( {169 - 20k} \right)}}{{100}} = 1\]\[ \Rightarrow 169 - 169 + 20k = 100\]169 is cancelled,\[ \Rightarrow 20k = 100\]Divide 100 by 20 we get,\[ \Rightarrow k = \dfrac{{100}}{{20}}\]\[ \Rightarrow k = 5\]

Thus the value of k is 5.

Note: Students first we will find the roots and then we will use the condition given. But students may get confused in the word reciprocal or they might forget to change the sign that is one root has plus sign \[\dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\] and other is having minus sign \[\dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\].

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Concept:

Let us consider the standard form of a quadratic equation, ax2 + bx + c =0

Let α and β be the two roots of the above quadratic equation.

The sum of the roots is given by: \({\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

The product of the roots is given by: \({\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

Calculation:

Let the roots of 5x2 + 26x + k = 0 are α and β

Given one root is reciprocal of the other So, α = 1/β

⇒ α.β = 1 ----(i)

Compare with general equation ax2 + bx + c = 0

a = 5, b = 26, c = k

According to the concept used

⇒ α.β = k/5 ----(ii)

From (i) and (ii), we get

k/5 = 1

∴The value of k is 5.

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