Answer to Question #125183 in Algebra for DZIMABI GEORGINA ATSUFUI

Question

Two functions f and g are defined on the set R of real numbers by
f(x) = ax2 + bx − 2 and g(x) = 3x + 2. If f(1) = 7 and f(2) = 10. Find:
(a) the values of a and b;
(b) the value(s) of x if f(x) = g(x) − 1.

Answer 1

$a) \ Given \ that \ f(x)=ax^2+bx-2 \\ f(1)=7 \ and \ f(2)=10\\ Substituting \ {x= 1} \ in \ f(x) \ and \ equating \ to \ 7, \ we \ get \\ a(1)^2+b(1)-2=7 \\\Rightarrow \ a+b =9 ...(1) \\$ $Substituting \ {x = 2} \ in \ f(x) \ and \ equating \ to \ 10, \ we \ get \\ a(2)^2+b(2)-2=10 \\ 4a+2b=12 \\ Dividing \ with \ {2}, \ we \ get \ 2a+b=6 ...(2)\\$$From \ the \ equation (1), \ { b = 9-a} ...(3)\\ Substituting \ {b=9-a} \ in \ {(2)},\ we \ get \\ 2a+9-a=6 \\ \Rightarrow \ {a+9=6} \\ \Rightarrow \ {a= 6-9=-3}\\ \therefore a =-3\\ Substituting \ {a=-3}, in \ (3), we \ get \ {b= 9-(-3)=12} \\ \therefore b=12.$ $b)\ Given \ that \ {f(x)= g(x)-1} \ and \ {g(x)=3x+2} \\ \Rightarrow \ ax^2+bx-2=3x+2\\ Substituting \ {a= -3} \ and \ {b=12} , \ we \ get \\$ $-3x^2+12x-2=3x+2-1\\ \Rightarrow \ -3x^2+12x-3x-2-2+1=0\\ \Rightarrow \ -3x^2+9x-3=0 \\ Dividing \ with \ {-3},\ we \ get \ {x^2-3x+1=0} \\ The \ roots \ of \ a \ quadratic \ are \ given \ by \\ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\$ $Substituting \ {a=1},\ {b=-3} \ and \ {c=1}, \ we \ get \\ x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(1)}}{2(1)}\\ =\frac{3\pm\sqrt{9-4}}{2}\\ =\frac{3\pm\sqrt{5}}{2}\\$ $\therefore { x_{1}=\frac{3+\sqrt{5}}{2}} \ and \ {x_{2}= \frac{3-\sqrt{5}}{2}}.$