Find the sum of 2x2 3x - 4, 8 - 3x, and -5x2 2. -3x2 - 2 -7x2 - 2 -3x2 6 -7x2 - 6x - 2

Mathematics

1 answer:

Law Incorporation [45]3 years ago

4 0

-3x to the second power - 2

I believe this is your answer.

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Answer:

$\sqrt{16 \times9 \times 3 \: \: equals \:4 \times 3 \times \sqrt{3that \: eauas \: 12 \sqrt{3} } }$

Step-by-step explanation:

Square root of 16 ×9×3= square root of 432

square root of 16 =4

square root of 9 =3

That will equal 4×3× square root of 3 which finally equals to 12 square root of 3.

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Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0