Question

Does the polynomial a4 + 4a2 + 5 have real zeroes?​

Answer 1

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✎In the aforementioned polynomial, let $a^2 = x.$

Now, the polynomial becomes,

$x^2 + 4x + 5$

Comparing with $ax^2 + bx + c,$

Here, $b^2 – 4ac = 4^2 – 4(1)(5) = 16 – 20 = -4$

So,$D = b^2 – 4ac < 0$

As the discriminant (D) is negative, the given polynomial does not have real roots

Answer 2

Answer:

Step-by-step explanation:

In such a case the equation has no real roots since square of any number cannot be negative. If it is positive then you can take square root and with plus minus sign you will get two roots. Note that if the values are -4 and +5 in the given question then there will be no real roots for the given equation

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