In a quadratic equation of the form ax²+bx+c=0 the discriminant is equal to (b²-4ac) if (b²-4ac) > 0 then the system has two real number solutions if (b²-4ac) =0 then the system has one real number solution if (b²-4ac) < 0 then the system has no real number solutions
we have y = x² and y = x + k
to resolve the system equate both equations x²=x+k-----------> x²-x-k=0 a=1 b=-1 c=-k find the disciminant (b²-4ac)=(-1)²-4*(1)*(-k)-----> 1+4k
Part 1)For which value of k does the system have no real number solutions? we know that if (b²-4ac) < 0 then the system has no real number solutions so (1+4k) < 0-------> 4k < -1 k< -0.25
the answer part 1) is for k < -0.25 the system has no real number solutions example if k=-2 -2 < -0.25-----> is ok the system has no real number solutions see the attached figure N 1
Part 2) For which value of k does the system have one real number solution? we know that if (b²-4ac) =0 then the system has one real number solution so (1+4k) = 0-------> k=-0.25
the answer part 2) is for k=-0.25 the system has one real number solution see the attached figure N 2
Part 3) For which value of k does the system have two real number solutions? we know that if (b²-4ac) > 0 then the system has two real number solutions so (1+4k) > 0------> 4k > -1 k > -0.25
the answer Part 3) is for k > -0.25 the system has two real number solutions example if k=2 2 > -0.25-----> is ok the system has two real number solutions see the attached figure N 3
