Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k$ A quadratics question.


  Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k.$


I recently asked a similar question, but this problem seems broader in scope.
Use the quadratic function, with a= 1, b= k, and c= -2.

There's three different outcomes for a quadratic:

${b^2-4ac}\gt 0 \rightarrow$ we get two solutions, which means we have two x-intercepts.

${b^2-4ac}=0 \rightarrow$ we get one solution, which means we have one x-intercept

${b^2-4ac}\lt 0 \rightarrow$ no real solutions, which means no x-intercepts. You can think of the parabola of the axis being above the x-axis if it opens upwards, or below the x-axis if it opens downwards.
for graph Y = x^2 + Kx -2 to cut x - axis y coordinate must be zero thus,

x^2 + Kx -2 = 0


Now for this equation for solution to be finitly 2 , b^2 - 4ac > 0.

here b = K , a = 1 and c = -2.

for every value of K , b^2 - 4ac will be greater than zero.

for example k = 0 => (0)^2 - 4(1)(-2) = 8

for k = -1 => (-1)^2 - 4(1)(-2) = 9

hence for all value of K it will be greater than zero so now we can find the solution of an equation.