Question

quadratic trinomial into factors 1+x-x²

Answer 1

Hey mate !

==> -x² + x + 1

a = -1

b = 1

c = 1

x= -b + √ b² - 4ac /2a

==> -1 + √ 1 - 4 (-1)(1) / 2(-1)

==> -1 + √ 1 + 4 /-2

==> -1 + √ 5/-2

==> 1 - √5/2


or

x = -b - √ b² - 4ac /2a

==> -1 - √ 1 - 4 (-1)(1) / 2(-1)

==> -1 - √ 1 + 4 /-2

==> -1 - √ 5/-2

==> 1+√5/2

Answer 2

$\underline{ \large{ \text{Answer}}} : \\ \\ \text{Now ,} \: 1 + \text{x} - { \text{x}}^{2} \\ \\ = 1 - ({ \text{x}}^{2} - \text{x}) \\ \\ = 1 - \{{ \text{x}}^{2} - 2( \text{x})( \frac{1}{2} ) + { (\frac{1}{2}) }^{2} - {( \frac{1}{2}) }^{2} \} \\ \\ = 1 - \{ {( \text{x} - \frac{1}{2} )}^{2} - \frac{1}{4} \} \\ \\ = 1 - ( \text{x} - \frac{1}{2})^{2} + \frac{1}{4} \\ \\ = (1 + \frac{1}{4} ) - {( \text{x} - \frac{1}{2} )}^{2} \\ \\ = (\frac{4 + 1}{4} ) - {( \text{x} - \frac{1}{2} )}^{2} \\ \\ = \frac{5}{4} - {( \text{x} - \frac{1}{2}) }^{2} \\ \\ = (\frac{ \sqrt{5} }{2})^{2} - {( \text{x} - \frac{1}{2} )}^{2} \\ \\ = ( \frac{ \sqrt{5} }{2} + \text{x} - \frac{1}{2} )( \frac{ \sqrt{5} }{2} - \text{x} + \frac{1}{2} ) \\ \\ = ( \frac{ \sqrt{5} - 1 }{2} - \text{x})( \frac{ \sqrt{5} + 1 }{2} - \text{x}) ,\\ \\ \text{which is the required factorization.} \\ \\ \underline{\text{Identity Rule}} : \\ \\ { \text{a}}^{2} - { \text{b}}^{2} = \text{(a + b)(a - b)} \\ \\ \bigstar \: \underline{ \large{ \text{MarkAsBrainliest}}} \: \bigstar$