# If one root of the polynomial \[k{{x}^{2}}-15x+18=0\] is 2, then find the value of k?

Last updated date: 25th Mar 2023

•

Total views: 306.3k

•

Views today: 6.83k

Answer

Verified

306.3k+ views

Hint:

We are given that x = 2 is the root of the given equation \[k{{x}^{2}}-15x+18=0\]. So, x = 2 will satisfy this equation. Substitute x = 2 in the equation and find the value of k.

Complete step-by-step answer:

Here, we are given that one of the roots of the polynomial \[k{{x}^{2}}-15x+18=0\] is 2 then we have to find the value of k.

First of all, we must know that roots or zeros of the given equation satisfy it or in other words, we can say that by substituting the value of roots in place of a variable in any equation, the equation becomes zero. Here variable is x, in terms of which that equation of polynomial is written.

Let us consider the polynomial given in this question.

\[k{{x}^{2}}-15x+18=0....\left( i \right)\]

We are given that 2 is the root of this polynomial. This means that x = 2 will satisfy this polynomial.

By substituting x = 2 in equation (i), we get,

\[k{{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0\]

By simplifying the above equation, we get,

\[\begin{align}

& 4k-30+18=0 \\

& \Rightarrow 4k-12=0 \\

\end{align}\]

By adding 12 on both sides of the above equation, we get,

\[4k-12+12=12\]

Or, \[4k=12\]

By dividing 4 on both sides of the equation, we get,

\[\dfrac{4k}{4}=\dfrac{12k}{4}\]

Or \[k=\dfrac{12}{4}\]

By simplifying the RHS of the above equation, we get,

\[\Rightarrow k=3\]

So, we get the value of k equal to 3.

Note:

In this question, we can cross-check our answer as follows:

Let us consider the polynomial \[k{{x}^{2}}-15x+18=0\]

By substituting k = 3 and x = 2 in the above polynomial, we get,

\[\left( 3 \right){{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0\]

By simplifying the above equation, we get,

\[3\left( 4 \right)-30+18=0\]

\[\Rightarrow 12-30+18=0\]

\[\Rightarrow 0=0\]

LHS = RHS

Since, LHS = RHS, therefore our answer is correct.

We are given that x = 2 is the root of the given equation \[k{{x}^{2}}-15x+18=0\]. So, x = 2 will satisfy this equation. Substitute x = 2 in the equation and find the value of k.

Complete step-by-step answer:

Here, we are given that one of the roots of the polynomial \[k{{x}^{2}}-15x+18=0\] is 2 then we have to find the value of k.

First of all, we must know that roots or zeros of the given equation satisfy it or in other words, we can say that by substituting the value of roots in place of a variable in any equation, the equation becomes zero. Here variable is x, in terms of which that equation of polynomial is written.

Let us consider the polynomial given in this question.

\[k{{x}^{2}}-15x+18=0....\left( i \right)\]

We are given that 2 is the root of this polynomial. This means that x = 2 will satisfy this polynomial.

By substituting x = 2 in equation (i), we get,

\[k{{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0\]

By simplifying the above equation, we get,

\[\begin{align}

& 4k-30+18=0 \\

& \Rightarrow 4k-12=0 \\

\end{align}\]

By adding 12 on both sides of the above equation, we get,

\[4k-12+12=12\]

Or, \[4k=12\]

By dividing 4 on both sides of the equation, we get,

\[\dfrac{4k}{4}=\dfrac{12k}{4}\]

Or \[k=\dfrac{12}{4}\]

By simplifying the RHS of the above equation, we get,

\[\Rightarrow k=3\]

So, we get the value of k equal to 3.

Note:

In this question, we can cross-check our answer as follows:

Let us consider the polynomial \[k{{x}^{2}}-15x+18=0\]

By substituting k = 3 and x = 2 in the above polynomial, we get,

\[\left( 3 \right){{\left( 2 \right)}^{2}}-15\left( 2 \right)+18=0\]

By simplifying the above equation, we get,

\[3\left( 4 \right)-30+18=0\]

\[\Rightarrow 12-30+18=0\]

\[\Rightarrow 0=0\]

LHS = RHS

Since, LHS = RHS, therefore our answer is correct.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

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

A Short Paragraph on our Country India