Question

x2 - 6x + 6 = 0 have
a) Real and Equal roots
b) Real roots
c) No Real roots
d) Real and Distinct roots​

Answer 1

Answer:

x²- 6x +6 = 0

∆ = b²- 4ac

= (-6)² - 4(1)6

= 36 - 24

∆ = 12

∆ > 0

So it has real and distinct roots...

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Answer 2

We have,

x² + 6x + 6 = 0

For the nature of roots, D = b² - 4ac

=> D = (6)² - 4(1)(6)

=> D = 36 - 24

=> D = 12

Thus D > 0

Thus the given equation has real and distinct roots.

Hence Proved

Hope this helps