Another Quadratic Equation Problem

nature of the roots is determined by the formula b^2 - 4ac. if b^2 - 4ac >0, then roots are real and distinct if b^2 - 4ac=0, then roots are real and equal if b^2 - 4ac<0, then roots are imaginary

b^2 - 4ac = 2^2 - 4(1)(-3) = 16 hence roots are real and distinct

The given quadratic equation is $x^2 + 2x - 3$

Finding solution to this equation is same as determining whether the $parabola$ of this equation intersects the $X$ axis or not .

If the parabola does $not$ intersect $X$ axis then there are $no$ $real$ roots.

If the parabola has its vertex on $X$ axis , then there are $same$ real roots.

If the parabola does intersect $X$ axis at $2$ $distinct$ points , then there are $2$ $different$ $real$ roots.

Since , leading coefficient of $x^2$ is $+1$ , the parabola opens upwards.

If $\Delta = b^2 - 4ac > 0$ then , the parabola does intersect X axis at 2 distinct points giving 2 different real roots.

Here, $\Delta = 2^2 -(4)(1)(-3) = 4 + 12 = 16 > 0$

Hence , there are 2 different real roots.