Conjugate of [tex]\sqrt{8} -\sqrt{9}[/tex] is equal to [tex]-3 -2\sqrt{2}[/tex].
" Conjugate is defined as the the irrational numbers which has property that sum and product number and its conjugate is a rational number. Sign of irrational and imaginary part is opposite that of original number."
According to the question,
Given number,
[tex]\sqrt{8} -\sqrt{9}[/tex]
Simplify it
[tex]2\sqrt{2}-3\\\\= -3 + 2\sqrt{2}[/tex]
Conjugate of [tex]-3 + 2\sqrt{2} = -3 -2\sqrt{2}[/tex] as [tex]2\sqrt{2}[/tex] is irrational part.
Sum of number and its conjugate
[tex](-3 + 2\sqrt{2} )+( -3 -2\sqrt{2}) = -6[/tex] is a rational number.
Product of number and its conjugate
[tex](-3 + 2\sqrt{2} )\times( -3 -2\sqrt{2}) \\\\=(-3)^{2} -(2\sqrt{2} )^{2} \\\\= 9-8\\\\=1[/tex]
[tex]1[/tex] is a rational number.
Hence, conjugate of [tex]\sqrt{8} -\sqrt{9}[/tex] is equal to [tex]-3 -2\sqrt{2}[/tex].
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