Nature of Roots of Quadratic Equation(With Methods & Examples)

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In real life, the quadratic equations help us in determining the area of space, the speed of a moving object, the value of profit gained on a product, and more. Even the path of a space rocket is described in terms of a quadratic equation. There are various methods of solving quadratic equations. 

Since the quadratic equations are degree-$2$ equations, they have $2$ roots or zeroes or solutions, which can be real numbers or imaginary numbers. You can determine the nature of the roots of quadratic equation without actually solving them.

Let’s understand how to find the nature of roots of quadratic equation without solving them.

Roots of Quadratic Equation

The values of a variable satisfying the given quadratic equation are called its roots. In other words, $x = x_{1}$ is a root of the quadratic equation $f\left(x \right)$, if $f\left(x_{1} \right) = 0$.

The real roots of an equation $f \left(x \right) = 0$ are the $x$-coordinates of the points where the curve $f \left(x \right)$ intersects the $x$-axis. These points are also known as the $x$-intercepts.

Nature of Roots of Quadratic Equation Based on Coefficients

Depending on the coefficients of a quadratic equation, the roots can be 

• one of the roots of the quadratic equation is zero and the other is $-\frac{b}{a}$ if $c = 0$
• both the roots are zero if $b = c = 0$
• The roots are reciprocal to each other if $a = c$

When $c = 0$

The standard form of a quadratic equation is $ax^{2} + bx + c = 0$

When $c = 0$, then the equation reduces to $ax^{2} + bx = 0$

$=>x \left(ax + b \right) = 0$

$=>x = 0$ or $ax + b = 0$

$=>x = 0$ or $x = -\frac{b}{a}$

When $b = c = 0$

The standard form of a quadratic equation is $ax^{2} + bx + c = 0$

When $b = c = 0$, then the equation reduces to $ax^{2} = 0$

$=>x^{2} = 0$

$=>x = 0$

Nature of Roots of Quadratic Equation Based on Discriminant

For a quadratic equation $ax^{2} + bx + c = 0$, the value $b^{2} – 4ac$ is called the discriminant of a quadratic equation and is denoted by either $\text{D}$ or $\triangle$.

The value of the discriminant can be zero $\left(0 \right)$, positive, or negative. Based on the value of a discriminant we can judge the nature of roots of quadratic equation without actually solving it.

• When $\text{D} = \triangle = b^{2} – 4ac = 0$, the two roots are real and equal
• When $\text{D} = \triangle = b^{2} – 4ac > 0$, the two roots are real and unequal
• When $\text{D} = \triangle = b^{2} – 4ac < 0$, the two roots are imaginary

Examples

Let’s consider some examples to understand the process of checking the nature of roots of a quadratic equation.

Ex 1: Find the nature of roots of a quadratic equation $2x^{2} + 8x + 7 = 0$.

Comparing $2x^{2} + 8x + 7$ with the standard form of quadratic equation $ax^{2} + bx + c$, we get the coefficients of the equation as $a = 2$, $b = 8$, and $c = 7$.

The discriminant of the equation $\text{D} = \triangle = b^{2} – 4ac = 8^{2} – 4 \times 2 \times 7 = 64 – 56 = 8$.

Since, $8 \gt 0$, therefore, the quadratic equation $2x^{2} + 8x + 7 = 0$ has two real unequal roots.

Let’s verify it by solving it.

$2x^{2} + 8x + 7 = 0$

By quadratic formula, $x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$

$=> x = \frac{-8 \pm \sqrt{8^{2} – 4 \times 2 \times 7}}{2 \times 2}$

$=> x = \frac{-8 \pm \sqrt{8}}{4}$

$=> x = \frac{-8 \pm 2\sqrt{2}}{4}$

$=> x = \frac{-4 \pm \sqrt{2}}{2}$

$=> x = -2 + \frac{\sqrt{2}}{2}$ and $x = -2 – \frac{\sqrt{2}}{2}$

Ex 2: Find the nature of roots of a quadratic equation $x^{2} + 6x + 9 = 0$.

Comparing $x^{2} + 6x + 9$ with the standard form of quadratic equation $ax^{2} + bx + c$, we get the coefficients of the equation as $a = 1$, $b = 6$, and $c = 9$.

The discriminant of the equation $\text{D} = \triangle = b^{2} – 4ac = 6^{2} – 4 \times 1 \times 9 = 36 – 36 = 0$.

Since, $0 = 0$, therefore, the quadratic equation $x^{2} + 6x + 9 = 0$ has two real equal roots.

Ex 3: Find the nature of roots of a quadratic equation $3x^{2} – 2x + 9 = 0$.

Comparing $3x^{2} – 2x + 9 = 0$ with the standard form of quadratic equation $ax^{2} + bx + c$, we get the coefficients of the equation as $a = 3$, $b = -2$, and $c = 9$.

The discriminant of the equation $\text{D} = \triangle = b^{2} – 4ac = \left(-2 \right)^{2} – 4 \times 3 \times 9 = 4 – 108 = -104$.

Since, $-104 \lt 0$, therefore, the quadratic equation $3x^{2} – 2x + 9 = 0$ has no real equal roots.

Note: When a quadratic equation has no real root, it means it has an imaginary root.

Nature of Roots of Quadratic Equation – Summary

The nature of roots of quadratic equation is summarized as follows.

• If the value of discriminant $\left(\text{D} \right) = 0$ i.e. $b^{2} – 4ac = 0$
• The quadratic equation will have equal roots i.e. $\alpha = \beta = -\frac{b}{2a}$
• If the value of discriminant $\left(\text{D} \right) \lt 0$ i.e. $b^{2} – 4ac \lt 0$
• The quadratic equation will have imaginary roots i.e $\alpha = \left(p + iq \right)$ and $\beta = \left(p – iq \right)$, ehere $iq$ is the imaginary part of a complex number if the value of $\left(\text{D} \right) \gt 0$ i.e. $b^{2} – 4ac \gt 0$
• The quadratic equation will have real roots if the value of discriminant $\left(\text{D} \right) \gt 0$ and $\text{ D }$ is a perfect square
• The quadratic equation will have rational roots if the value of $\left(\text{D} \right) \gt 0$ and $\text{ D }$ is not a perfect square
• The quadratic equation will have irrational roots i.e. $\alpha = \left(p + \sqrt{q} \right)$ and $\beta = \left(p – \sqrt{q} \right)$ if the value of $\left(\text{D} \right) \gt 0$, $\text{D}$ is a perfect square, $a = 1$ and $b$ and $c$ are integers
• The quadratic equation will have integral roots if the value of $\left(\text{D} \right) \gt 0$, $\text{D}$ is a perfect square, $a = 1$ and $b$ and $c$ are integers

Condition for Common Root or Roots of Quadratic Equations

Let the two quadratic equations are $a_{1}x^{2} + b_{1}x + c_{1} = 0$ and $a_{2}x^{2} + b_{2}x + c_{2} = 0$

Now we are going to find the condition that the above quadratic equations may have a common root.

Let $\alpha$ be the common root of the equations $a_{1}x^{2} + b_{1}x + c_{1} = 0$ and $a_{2}x^{2} + b_{2}x + c_{2} = 0$. Then,

$a_{1} \alpha^{2} + b_{1} \alpha + c_{1} = 0$

$a_{2} \alpha^{2} + b_{2} \alpha + c_{2} = 0$

Now, solving the equations $a_{1} \alpha^{2} + b_{1} \alpha + c_{1} = 0$, $a_{2} \alpha^{2} + b_{2} \alpha + c_{2} = 0$ by cross-multiplication, we get

$\frac{\alpha^{2}}{b_{1}c_{2} – b_{2}c_{1}} = \frac{α}{c_{1}a_{2} – c_{2}a_{1}} = \frac{1}{a_{1}b_{2} – a_{2}b_{1}}$

$=> \alpha = \frac{b_{1}c_{2} – b_{2}c_{1}}{c_{1}a_{2} – c_{2}a_{1}}$, (From first two)

Or, $\alpha = \frac{c_{1}a_{2} – c_{2}a_{1}}{a_{1}b_{2} – a_{2}b_{1}}$, (From $2^{nd}$ and $3^{rd}$)

$=>\frac{b_{1}c_{2} – b_{2}c_{1}}{c_{1}a_{2} – c_{2}a_{1}} = \frac{c_{1}a_{2} – c_{2}a_{1}}{a_{1}b_{2} – a_{2}b_{1}}$

$=> \left(c_{1}a_{2} – c_{2}a_{1} \right)^{2} = \left(b_{1}c_{2} – b_{2}c_{1} \right) \left(a_{1}b_{2} – a_{2}b_{1} \right)$, which is the required condition for one root to be common of two quadratic equations.

Examples

Ex 1: For what value of $k$, both the quadratic equations $6x^{2} – 17x + 12 = 0$ and $3x^{2} – 2x + k = 0$ will have a common root.

If one of the root of quadratic equations $a_{1}x^{2} + b_{1}x + c_{1} = 0$ and $a_{2}x^{2} + b_{2}x + c_{2} = 0$ is common then 

$\left(a_{1}b_{2} – a_{2}b_{1} \right) \left(b_{1}c_{2} – b_{2}c_{1} \right) = \left(a_{2}c_{1} – a_{1}c_{2} \right)^{2}$ —————————(1)

Form the given quadratic Equations, $a_{1} = 6$, $b_{1} = -17$, $c_{1} = 12$, $a_{2} = 3$, $b_{2} = -2$ and $c_{2} = k$.

On substituting these values in equation (1), we will get:

$\left( \left(6 \times \left(-2 \right) \right) – \left(3\times \left(-17 \right) \right) \right) \times \left(-17k – \left(-2 \times 12 \right) \right) = \left( 3 \times 12 – 6k \right)^{2}$

$=>-663k + 936 = 1296 + 36k^{2} – 432k$

$=>36k^{2} + 231k + 360 = 0$

$=>12k^{2} + 125k + 120 = 0$

$=> \left(4k + 15 \right) \left(3k + 8 \right) = 0$

Therefore, the values of $k$ are $-\frac{15}{4}$, $-\frac{8}{3}$.

Ex 2: Find the values of $k$ such that the quadratic equations $x^{2} – 11x + k = 0$ and $x^{2} – 14x + 2k = 0$ have a common factor.

Let $\left(x – \alpha \right)$ be the common factor of quadratic equations $x^{2} – 11x + k = 0$ and $x^{2} – 14x + 2k  = 0$ then $x = \alpha$ will satisfy the given quadratic equations.

Therefore, $\alpha^{2} – 11 \alpha + k = 0$ ———————- (1)

And, $\alpha^{2} – 14 \alpha + 2k = 0$  ——————— (2)

On Solving Equation (1) and Equation (2) we will get:

$\frac{\alpha^{2}}{-22k + 14k} = -\frac{\alpha}{2k – k} = \frac{1}{-14 + 11}$

Therefore, $α^{2} = \frac{-22k + 14k}{-3} = \frac{8k}{3}$ ——————— (3)

And, $\alpha = \frac{2k – k}{-14 + 11} = \frac{k}{3} ———————– (4)

On Equating Equation (3) and Equation (4):

$\frac{8k}{3} = \left(\frac{k}{3} \right)^{2}

Therefore, the value of $k = 24$.

Practice Problems

1. Find the nature of roots for the following quadratic equations
   • $2x^{2} + 5x – 3$
   • $x^{2} – 7x – 2$
   • $4x^{2} + x + 6$
   • $2x^{2} – 2x + 1$
2. For what value(s) of $k$, do the following quadratic equations have equal real roots
   • $kx^{2} + x – 2$
   • $2x^{2} + 4x + 8$
3. For what value(s) of $k$, do the following quadratic equations have no real roots
   • $x^{2} + kx – 5$
   • kx^{2} + 2x + 8$

FAQs

Conclusion

Since the quadratic equations are degree-$2$ equations, they have $2$ roots or zeroes or solutions, which can be real numbers or imaginary numbers. Using the value of discriminant one can verify the zeroes or roots of a quadratic equation without actually solving it!

Recommended Reading

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