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Mathematics

If 2\sqrt{2} is a root of the equation kx2+2x4=0kx^2 + \sqrt{2}x - 4 = 0, find the value of k.

Quadratic Equations

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Answer

Since 2\sqrt{2} is a root of the equation kx2+2x4=0kx^2 + \sqrt{2}x - 4 = 0, x=2x = \sqrt{2} satisfies the given equation.

Substituting x=2x = \sqrt{2} in the given equation:

k(2)2+2×24=02k+24=02k2=02k=2k=1k ( \sqrt{2} )^2 + \sqrt{2} \times \sqrt{2} - 4 = 0 \\[0.5em] \Rightarrow 2k + 2 - 4 = 0 \\[0.5em] \Rightarrow 2k - 2 = 0 \\[0.5em] \Rightarrow 2k = 2 \\[0.5em] \Rightarrow k = 1

Hence, the value of k is 1

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