Question

.12th pls do it pls marking him brailiest​

Answer 1

Answer:

$11. \: {(x + 2)}^{3} = {x}^{3} - 4$

$x {}^{3} + {6x}^{2} + 12x + 2 {}^{3} = x {}^{3} - 4$

$6x {}^{2} + 12x + 12 = 0$

$hence \: it \: is \: prooved$

$12. \: (k - 1) {x}^{2} - 30 x + 25 = 0$

$(k - 1) {x}^{2} - 2 \times 5 \times 3x + {5}^{2}$

$k - 1 = 3$

$k = 4$

$= (3x - {5)}^{2}$

$13. \: x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$a = 1.b = 4.c = 5$

$x = \frac{ - 4± \sqrt{ {4}^{2} - 4 \times 1 \times 5 } }{2}$

$x = \frac{ - 4± \sqrt{16 - 20} }{2}$

$as \: the \: formula \: if \: {b}^{2} - 4ac \: is \: negative \: the \: roots \: \alpha \: and \: \beta \: are \: imaginary \: and \: unequal.$

$x = \frac{ - 4± \sqrt{ - 4} }{2}$

$so \: the \: equation \: has \: no \: real \: roots$

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