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Last modified on August 3rd, 2023

Some quadratic equations have no real solution. There are different ways by which we can identify whether a quadratic equation can have a solution or not.

The most widely used method to identify whether a quadratic equation has a solution is by looking at the value of the discriminant.

Suppose the value of the discriminant is less than 0 (b^{2} – 4ac < 0) in the quadratic equation ax2 + bx + c, the equation will have no real solution. In another way, if b^{2} < 4ac, the equation will give complex roots with a negative sign within the square root.

Let us consider the quadratic equation x^{2} – 4x + 5 = 0, here a = 1, b = -4, c = 5

Using the quadratic formula x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ and plugging in the values of a, b, and c to find its roots, we get,

= ${\dfrac{-\left( -4\right) \pm \sqrt{\left( -4\right) ^{2}-4\times 1\times 5}}{2\left( 1\right) }}$

= ${\dfrac{4\pm \sqrt{16-20}}{2}}$

= ${\dfrac{4\pm \sqrt{-4}}{2}}$

As we know, the square root of a negative number is not a real number. Thus, this quadratic equation has no real roots or solutions. However the equation have two complex solutions {2 + i, 2 – i}

**Find if the quadratic equation x ^{2} + 2x + 5 = 0 has real solutions.**

Solution:

Comparing this equation with the standard form of the equation ax^{2} + bx + c, here a = 1, b = 2, c = 5

Plugging in the values in the quadratic equation, we get

= ${\dfrac{-2\pm \sqrt{\left( 2\right) ^{2}-4\times 1\times 5}}{2\times 1}}$

= ${\dfrac{-2\pm \sqrt{4-20}}{2}}$

Â = ${\dfrac{-2\pm \sqrt{-16}}{2}}$

As we know, the square root of a negative number is not a real number. Thus, this quadratic equation has no real roots or solutions. However the equation have two complex solutions {2i – i, -2i – i}

Another exciting way to identify whether a quadratic equation has a real solution is by looking at the graph. If the graph does not touch the x-axis, it will have no real solution.

Let us consider the same quadratic equation x^{2} – 4x + 5 = 0 to conclude. Suppose we plot the graph using the coordinate points found manually or a graphing calculator. In that case, we can indeed find that the parabola does not touch the x-axis.

So, the concept indeed works. Letâ€™s try out another such example.

**Find if the quadratic equation -x ^{2} – 3x – 10 = 0 has real roots.**

Solution:

Graphing the given quadratic equation using a plotter, we get the parabola given below:

Since this parabola does not touch the x-axis, we can conclude that its corresponding quadratic equation has no real solutions. Also, note that a quadratic equation with no real solution could have a graph below the x-axis.

If we solve the given equation using the quadratic formula, we will find two complex solutions {1 + i, 1 – i}

Another simple way to determine if the given quadratic equation has real roots or solution is by looking at the coefficients, given the equation is written in standard form.

If the middle-term coefficient is zero (b = 0) and both â€˜aâ€™ and â€˜bâ€™ share the same sign (positive or negative), then the equation will have no real solution. After simplification, we will find that the solutions are the positive and negative square roots of ${-\dfrac{c}{a}}$

There can be two possible cases:

- If c and a are both positive, then ${\dfrac{c}{a}}$ is positive, and ${-\dfrac{c}{a}}$ is negative
- If c and a are both negative, then ${\dfrac{c}{a}}$ is positive, and ${-\dfrac{c}{a}}$ is negative

In either case, we need to do the square root of a negative number, which will give us two complex solutions.

For example in the quadratic equation -2x^{2} â€“ 16 = 0, a = -2, b = 0, c = -16. Since b = 0 and both â€˜aâ€™ and â€˜câ€™ have the same sign, there will be no real solution to this equation. However the equation have two complex solutions { ${2\sqrt{2}i}$, ${-2\sqrt{2}i}$}

**Find if the quadratic equation x ^{2} + 22 = 0 has real roots.**

Solution:

Comparing the given equation with the quadratic equation in the standard form ax^{2} + bx + c, we get a = 1, b = 0, c = 22 Since b = 0 and both â€˜aâ€™ and â€˜câ€™ have the same sign, there will be no real solution to this equation. However the equation have two complex solutions {${\sqrt{22}i}$, ${-\sqrt{22}i}$}

Last modified on August 3rd, 2023