Question

prove that the roots of the equcation
x²+(2a- 1) x+a2 = 0 are real if a leqslant 1/4

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Answer 1

Step-by-step explanation:

check D

${b}^{2} - 4ac \geqslant 0 \\ (2a - 1) {}^{2} - 4 {a}^{2} \geqslant 0 \\ (2a - 1 - 2a)(2a - 1 + 2a) \geqslant 0 \\ - (4a - 1) \geqslant 0 \\4a - 1 \leqslant 0 \\ a \leqslant \frac{1}{4}$