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# How do you determine whether there are two, one or no real solutions given the graph of a quadratics function does not have an $x$ -intercept?

Last updated date: 01st Mar 2024
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Hint:When we say there is no $x$ -intercept it means that it does not cross the $x$ -axis. In other words, we can say that if a line has no $x$ -intercept, then it never intersects $x$ -axis which means that it is parallel to $y$ -axis. So, we can conclude that this is a vertical line and its slope is undefined. Therefore, the possibility of two solutions is definitely ruled out.
Here, in this question students have to determine whether there are two, one or no real solutions for the graph of a quadratic function which does not have a $x$ -intercept.
Let us assume that we include a point of coincidence i.e., the vertex coincides with the $x$ -axis, then the plot doesn’t cross the $x$-axis nor does any point on the curve coincide with it. In such an assumption, there are no real solutions.
The $x$ -axis is composed of all those points for which $f\left( x \right)$ is equal to zero.
If the graph of $f\left( x \right)$ doesn’t have an $x$ -intercept then, it means that it has no real solutions or points for which $f\left( x \right) = 0$.
Hence, we conclude that if the graph of a quadratic function does not have an $x$ -intercept then, it has no real solutions or roots.
Note: Students can also check whether a quadratic equation has two, one or no real solutions by using a quadratic formula. In the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, the expression$\sqrt {{b^2} - 4ac}$ is called as discriminant and is often denoted by $D$. If $D$ is positive or greater than zero, then the two roots of the equation are real. If $D$ is zero, then roots are real but if $D$ is negative or less than zero, then roots are not real.