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In the Following, Find the Value of K for Which the Given Value is a Solution of the Given Equation: `Kx^2+Sqrt2x-4=0`, `X=Sqrt2` - Mathematics

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Question

In the following, find the value of k for which the given value is a solution of the given equation:

`kx^2+sqrt2x-4=0`, `x=sqrt2`

Solution

We are given here that,

`kx^2+sqrt2x-4=0`, `x=sqrt2`

Now, as we know that `x=sqrt2` is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting `x=sqrt2` in the above equation gives us,

`kx^2+sqrt2x-4=0`

`k(sqrt2)^2+sqrt2(sqrt2)-4=0`

2k + 2 - 4 = 0

2k - 2 = 0

2k = 2

`k=2/2`

k = 1

Hence, the value of k = 1.

  Is there an error in this question or solution?
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APPEARS IN

 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.1 | Q: 3.3 | Page no. 4
 RD Sharma Solution for Class 10 Maths (2018 (Latest))
Chapter 4: Quadratic Equations
Ex. 4.1 | Q: 3.3 | Page no. 4
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In the Following, Find the Value of K for Which the Given Value is a Solution of the Given Equation: `Kx^2+Sqrt2x-4=0`, `X=Sqrt2` Concept: Quadratic Equations.
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