If one zero of the polynomials \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\] is the reciprocal of the other, then the value of \[a\] is
Answer
382.5k+ views
Hint: In the given polynomial \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\] the degree (highest power) is two. So, the polynomial has two zeros or roots.
Let \[p\] be one of the zero of the polynomials \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\]. Then the other zero is \[\dfrac{1}{p}\] since one of the zero is the reciprocal of the other.
We know that for the quadratic polynomial \[a{x^2} + bx + c = 0\], the sum of the roots is \[ - \dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\].
Now consider the product of the zeroes or roots
i.e. \[p \times \dfrac{1}{p} = \dfrac{{6a}}{{{a^2} + 9}}\]
\[
1 = \dfrac{{6a}}{{{a^2} + 9}} \\
{a^2} + 9 = 6a \\
{a^2} - 6a + 9 = 0 \\
\]
By solving \[{a^2} - 6a + 9 = 0\] we get
\[
{a^2} - 3a - 3a + 9 = 0 \\
a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0 \\
\left( {a - 3} \right)\left( {a - 3} \right) = 0 \\
{\left( {a - 3} \right)^2} = 0 \\
\therefore a = 3 \\
\]
Therefore, the value of \[a\] is \[3\].
Note: The given polynomial is quadratic polynomial since the degree (highest power) is two. In this problem we can also consider the sum of the zeroes or roots but it is a lengthy process. So, we considered the product of the zeroes or roots.
Let \[p\] be one of the zero of the polynomials \[\left( {{a^2} + 9} \right){x^2} + 13x + 6a\]. Then the other zero is \[\dfrac{1}{p}\] since one of the zero is the reciprocal of the other.
We know that for the quadratic polynomial \[a{x^2} + bx + c = 0\], the sum of the roots is \[ - \dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\].
Now consider the product of the zeroes or roots
i.e. \[p \times \dfrac{1}{p} = \dfrac{{6a}}{{{a^2} + 9}}\]
\[
1 = \dfrac{{6a}}{{{a^2} + 9}} \\
{a^2} + 9 = 6a \\
{a^2} - 6a + 9 = 0 \\
\]
By solving \[{a^2} - 6a + 9 = 0\] we get
\[
{a^2} - 3a - 3a + 9 = 0 \\
a\left( {a - 3} \right) - 3\left( {a - 3} \right) = 0 \\
\left( {a - 3} \right)\left( {a - 3} \right) = 0 \\
{\left( {a - 3} \right)^2} = 0 \\
\therefore a = 3 \\
\]
Therefore, the value of \[a\] is \[3\].
Note: The given polynomial is quadratic polynomial since the degree (highest power) is two. In this problem we can also consider the sum of the zeroes or roots but it is a lengthy process. So, we considered the product of the zeroes or roots.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
