Show that 9x²+6x+4 has no real zeroes
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Given polynomial = 9x² + 6x + 4
To show that 9x²+6x+4 = 0 has no real solutions.
A quadratic equation has no real solution if the value of it's discriminant is negative.
If equation is of the form ax² + bx + c = 0
D = b² - 4ac
Comparing the equations , a = 9, b = 6, c = 4
D = 6² - 4×9×4
D = 36 - 144
D = -108
D < 0
Since discriminant is negative, it implies this equation has no real roots.
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