Question

Prove that x² +4x+5 has no real zero.
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Answer 1

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D=b²-4ac. D= -4. So, as the D is negative, The roots will be imaginary and hence there will be no real roots!

Answer 2

The equation has no real zeroes.

Given: The equation = $x^{2} +4x+5=0$.

To Prove: This equation has no real zeroes.

Solution:

If a equation has no real zeroes then its D < 0.

In equation $ax^{2} +bx+c=0$.

D = $b^{2} -4ac$

In equation $x^{2} +4x+5=0$.

a = 1, b = 4 and c = 5.

D = $4^{2} -4(1)(5)$

D = 16 - 20

D = -4

As D < 0, it is proved that the equation x² +4x+5 has no real zeroes.

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