Question

4/ Let a,b, and be constants (with a 0), and consider the system ly= ax² + bxtc 1=k For which value of k (in terms of a, b, and c) will the system have exactly one solution? What is that solution? What is the relationship between the solution you've found and the graph of y=ar? +bx+e?

Answer 1

Given a b and c are constants (with a to and the system is sy = ax?+bx+e у — К. Know that any quodratic equation has Solution if the quadratic equation's discriminant one cwill be Zero. 2 Discriminant = Coefficient of a - 4x coefficientot ee? Constant term. i.e For (1) Discriminant = 2 b - 4.A.C. Also for the Solution of the system Equate () & (ii) ie ax²+bx+c =k 7 ax²+bx+ck=0 Now for exactly one solution Discriminant of (3) must be zero i e B - 4.a. (cak= 0 - b 4a (C-K) » 4a (C-k) = 6² C-K = b 4a 2. 2 - c- 6² ( adding 'k' both sides 40 & subtracting & both sides) K = C- 2 be 4a

Hence For K = C- b (in terms of a b&c 40 the system will have exactly one solution. Now and part What is that solutionn: We have: ax² + 6x + C-k = 0 (by equation c By, Shridharacharya, (3) -b √ Discriminant x = b= Coefficient of a za a = Coefficient of x² -b & To -> x = For exactly one solution discriminant of (3) will be zero zero) x = 2 za -b Hence the solution is x = -b 20 Now third part: What is the relationship between the solution you have found and the graph of yzax? +bx+c ? solution - The Relationship between the Solution that I found and the graph of y= ax²+bx+ c is : At this Solution of the value of a the height of y= ax²+bx+c will be same from the x-axis. also one can get the value of k easily by putting the Value of x ie solution ing=ax² +bate. & y=k