Question

12. Solve the quadratic equations,
x2 + 3x + 9 = 0​

Answer 1

Answer:

Therefore,

\begin{gathered}:\implies\sf 10y + x = (10x + y) - 54\\ \\ \\ :\implies\sf 10y - y + x - 10x = - 54\\ \\ \\ :\implies\sf 9y - 9x = - 54\\ \\ \\ :\implies\sf 9(y - x) = - 54\\ \\ \\:\implies\sf y - x = \cancel{\dfrac{-54}{9}}\\ \\ \\ :\implies\sf y - x = - 6\\ \\ \\ \dag\;{\underline{\frak{Substituting\:value\:of\:'x'\:from\:eq\:(1),}}}\\ \\ \\ :\implies\sf y - 4y = -6\\ \\ \\ :\implies\sf -3y = -6\\ \\ \\ :\implies\sf y = \cancel{\dfrac{-6}{-3}}\\ \\ \\:\implies{\underline{\boxed{\frak{\purple{y = 2}}}}}\;\bigstar\\ \\\end{gathered}

:⟹10y+x=(10x+y)−54

:⟹10y−y+x−10x=−54

:⟹9y−9x=−54

:⟹9(y−x)=−54

:⟹y−x=

9

−54

:⟹y−x=−6

†

Substitutingvalueof

′

x

′

fromeq(1),

:⟹y−4y=−6

:⟹−3y=−6

:⟹y=

−3

−6

:⟹

y=2

★

Answer 2

Step-by-step explanation:

x² + 3x + 9 = 0

The above equation is in the form

ax² + bx + c = 0

where a = 1 , b = 3 , c = 9

x = -b ± √b² - 4ac

_____________

2a

x = -3 ± √3² - 4 × 1 × 9

______________

2

x = -3 ± √ 9 + 36

___________

2

= -3 ± √ -27

________

2

= -3 ± √ -1 × 27

____________

2

= -3 ± ( √ 27 × √ -1 )

______________

2

= -3 ± √ 27 i

_________

2

= -3 ± 3√3i

________

2

x = 3 ± 3√3i

______

2