# If one root of the quadratic equation \[2{{x}^{2}}+px-4=0\] is ‘2’, then find the value of ‘p’ will be:

a)-3

b)-2

c)2

d3

Last updated date: 23rd Mar 2023

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Answer

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Hint: To solve the question, we have to apply the sum and product of roots of quadratic equations formulae and calculate the unknown values.

Complete step-by-step answer:

Given

The quadratic equation is \[2{{x}^{2}}+px-4=0\].

The root of the quadratic equation is given as 2.

Let the other root of the quadratic equation be y.

We know that the formula of the sum of roots of a general quadratic equation \[a{{x}^{2}}+bx+c=0\] is equal to \[\dfrac{-b}{a}\] and the formula of the product of roots is equal to \[\dfrac{c}{a}\].

By comparing the general quadratic equation with our quadratic equation, we get

a = 2, b = p, c = -4.

The sum of roots of the equation will become\[\dfrac{-p}{2}\]

\[2+y=\dfrac{-p}{2}\]

\[\Rightarrow y=\dfrac{-p}{2}-2\] ……\[(1)\]

The product of roots of the equation\[=\dfrac{-4}{2}=-2\]

\[2y=-2\]

\[\Rightarrow y=-1\]

By substituting the above value in equation (1), we get

\[-1=\dfrac{-p}{2}-2\]

\[1=\dfrac{-p}{2}\]

\[\Rightarrow p=-2\]

The value of p is equal to -2

Hence, the option (b) is the right choice.

Note: The possibility of mistake can be found at the application of the formulae for the sum and the product of the roots of the quadratic equation. The possibility of mistake is calculations as mistakes are possible because of various positive and negative values. The alternative method is by substituting x = 2 in the given quadratic equation and by solving the equation we can find the value of p. The other alternative method of solving can be using the hit-trial method, substitute the options in the given quadratic equation and check whether the root x = 2 can satisfy or not.

Complete step-by-step answer:

Given

The quadratic equation is \[2{{x}^{2}}+px-4=0\].

The root of the quadratic equation is given as 2.

Let the other root of the quadratic equation be y.

We know that the formula of the sum of roots of a general quadratic equation \[a{{x}^{2}}+bx+c=0\] is equal to \[\dfrac{-b}{a}\] and the formula of the product of roots is equal to \[\dfrac{c}{a}\].

By comparing the general quadratic equation with our quadratic equation, we get

a = 2, b = p, c = -4.

The sum of roots of the equation will become\[\dfrac{-p}{2}\]

\[2+y=\dfrac{-p}{2}\]

\[\Rightarrow y=\dfrac{-p}{2}-2\] ……\[(1)\]

The product of roots of the equation\[=\dfrac{-4}{2}=-2\]

\[2y=-2\]

\[\Rightarrow y=-1\]

By substituting the above value in equation (1), we get

\[-1=\dfrac{-p}{2}-2\]

\[1=\dfrac{-p}{2}\]

\[\Rightarrow p=-2\]

The value of p is equal to -2

Hence, the option (b) is the right choice.

Note: The possibility of mistake can be found at the application of the formulae for the sum and the product of the roots of the quadratic equation. The possibility of mistake is calculations as mistakes are possible because of various positive and negative values. The alternative method is by substituting x = 2 in the given quadratic equation and by solving the equation we can find the value of p. The other alternative method of solving can be using the hit-trial method, substitute the options in the given quadratic equation and check whether the root x = 2 can satisfy or not.

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