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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?

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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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.

Complete step by step solution:
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.
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