Question

Which of the following are solutions to the equation below?

Answer 1

$\\ \rm\bull\rightarrowtail x^2-3x+27=6x+7$

$\\ \rm\bull\rightarrowtail x^2-3x-6x+27-7=0$

$\\ \rm\bull\rightarrowtail x^2-9x+20=0$

$\\ \rm\bull\rightarrowtail x^2-5x-4x+20=0$

$\\ \rm\bull\rightarrowtail x(x-5)-4(x-5)=0$

$\\ \rm\bull\rightarrowtail (x-4)(x-5)=0$

$\\ \rm\bull\rightarrowtail x=4,5$

Answer 2

♪Answer :

»x² - 3x + 27 = 6x + 7

• ax² + bx + c = 0

»x² - 3x + 27 - 6x - 7 = 0

»x² - 9x + 20 = 0

»(x - 4)(x - 5) = 0

»(x - 4) = 0 V (x - 5) = 0

• For $\tiny x_1$

»x - 4 = 0

»x = 0 + 4

»x = 4

• For $\tiny x_2$

»x - 5 = 0

»x = 0 + 5

»x = 5

• x = {x| 4 or 5, x€R}