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The equation 2x^2 + 2kx + (k + 2) = 0, where k is a constant, has two distinct real roots. Show that k satisfies k^2 – 2k – 4 > 0

Two distinct real roots means that we can use b^2-4ac>0 relationship for any ax^2+bx+c equation. Apply the above gives, 4k^2 - 42(k+2)>0 Simplifying gives, k^2 - 2k -4 >0

Answered by Andreas T. • Maths tutor

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The equation 2x^2 + 2kx + (k + 2) = 0, where k is a constant, has two distinct real roots. Show that k satisfies k^2 – 2k – 4 > 0

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