[tex]4(x - 5) {}^{2} - 9 \\ = 4( {x}^{2} - 10x + 25) - 9 \\ = 4x {}^{2} - 40x + 100 - 9 \\ = 4x {}^{2} - 40x + 91 \\ = f(x)[/tex]
[tex](2x - 13)(2x - 7) \\ = 4x {}^{2} - 14x - 26x + 91 \\ = 4x {}^{2} - 40x + 91 \\ = f(x)[/tex]
a)
[tex]f(1) = 4(1) {}^{2} - 40(1) + 91 \\ = 4 - 40 + 91 \\ = 55[/tex]
b)
[tex]f(x) = 0 \\ (2x - 13)(2x - 7) = 0 \\ 2x - 13 = 0 \\ 2x = 13 \\ x = \frac{13}{2} \\ ou \\ 2x - 7 = 0 \\ 2x = 7 \\ x = \frac{7}{2} [/tex]
c)
[tex]f(x) = 91 \\ 4 {x}^{2} - 40x + 91 = 91 \\ 4x {}^{2} - 40x = 91 - 91 \\ 4x {}^{2} - 40x = 0 \\4 x( x - 10) = 0 \\ x = 0 \: \\ ou \\ x - 10 = 0 \\ x = 10[/tex]
donc les antécédents de 91 sont 0 et 10
d)
[tex]f(x) = 7 \\ 4 {x}^{2} - 40x + 91 = 7 \\ 4x {}^{2} - 40x + 91 - 7 = 0 \\ 4 {x}^{2} - 40x + 84 = 0 \\ 4( {x}^{2} - 10x + 21) = 0 \\ {x}^{2} - 10x + 21 = 0 \\ x = 3 \: ou \: x = 7[/tex]
