Answer
Hint: The nature of the roots of a quadratic equation can be determined from the value of the discriminant. The roots of an equation are the points at which the curve meets the x-axis. So, we can find the nature of the root and thus predict the graph.
Complete step by step answer:
We know that the nature of the roots of a quadratic equation can be found from its discriminant. If the discriminant is greater than zero, the equation will have two real and distinct roots.
If the discriminant is zero, the equation will have a real root. If the discriminant is less than zero, the equation will have no real roots, it will have 2 complex roots.
Graphically, the roots of an equation can be defined as the points where the curve of the equation meets the x-axis.
So, if an equation has 2 roots, then the curve meets the x-axis at two points. Its graph is given by,
If the equation has no real then the curve does not meet the x-axis and its graph is given by,
And if the equation has only 1 root, the graph meets the x-axis at only one point. Its graph is given by,
We are given that the quadratic equation has discriminant zero. So, the equation has only one root. So, the graph of the equation will touch the x-axis only once.
Therefore, the correct answer is option B.
Note: Discriminant of quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $D = {b^2} - 4ac$.
If $D > 0$, the equation has 2 real and distinct roots. They are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
If $D = 0$, the equation has a real root, which is given by, $x = \dfrac{{ - b}}{{2a}}$
If $D < 0$, the equation has complex roots.
The quadratic formula
The discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac. The discriminant tells us whether there are two solutions, one solution, or no solutions.
The quadratic formula
Examining the roots of a quadratic equation means to see the type of its roots i.e., whether they are real or imaginary, rational or irrational, equal or unequal.
The nature of the roots of a quadratic equation depends entirely on the value of its discriminant b\(^{2}\) - 4ac.
In a quadratic equation ax\(^{2}\) + bx + c = 0, a ≠ 0 the coefficients a, b and c are real. We know, the roots (solution) of the equation ax\(^{2}\) + bx + c = 0 are given by x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).
1. If b\(^{2}\) - 4ac = 0 then the roots will be x = \(\frac{-b ± 0}{2a}\) = \(\frac{-b - 0}{2a}\), \(\frac{-b + 0}{2a}\) = \(\frac{-b}{2a}\), \(\frac{-b}{2a}\).
Clearly, \(\frac{-b}{2a}\) is a real number because b and a are real.
Thus, the roots of the equation ax\(^{2}\) + bx + c = 0 are real and equal if b\(^{2}\) – 4ac = 0.
2.
If b\(^{2}\) - 4ac > 0 then \(\sqrt{b^{2} - 4ac}\) will be real and non-zero. As a result, the roots of the equation ax\(^{2}\) + bx + c = 0 will be real and unequal (distinct) if b\(^{2}\) - 4ac > 0.
3. If b\(^{2}\) - 4ac < 0, then \(\sqrt{b^{2} - 4ac}\) will not be real because \((\sqrt{b^{2} - 4ac})^{2}\) = b\(^{2}\) - 4ac < 0 and square of a real number always positive.
Thus, the roots of the equation ax\(^{2}\) + bx + c = 0 are not real if b\(^{2}\) - 4ac < 0.
As the value of b\(^{2}\) - 4ac determines the nature of roots (solution), b\(^{2}\) - 4ac is called the discriminant of the quadratic equation.
Definition of discriminant: For the quadratic equation ax\(^{2}\) + bx + c =0, a ≠ 0; the expression b\(^{2}\) - 4ac is called discriminant and is, in general, denoted by the letter ‘D’.
Thus, discriminant D = b\(^{2}\) - 4ac
Note:
Discriminant of ax\(^{2}\) + bx + c = 0 | Nature of roots of ax\(^{2}\) + bx + c = 0 | Value of the roots of ax\(^{2}\) + bx + c = 0 |
b\(^{2}\) - 4ac = 0 | Real and equal | - \(\frac{b}{2a}\), -\(\frac{b}{2a}\) |
b\(^{2}\) - 4ac > 0 | Real and unequal | \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) |
b\(^{2}\) - 4ac < 0 | Not real | No real value |
When a quadratic equation has two real and equal roots we say that the equation has only one real solution.
Solved examples to examine the nature of roots of a quadratic equation:
1. Prove that the equation 3x\(^{2}\) + 4x + 6 = 0 has no real roots.
Solution:
Here, a = 3, b = 4, c = 6.
So, the discriminant = b\(^{2}\) - 4ac
= 4\(^{2}\) - 4 ∙ 3 ∙ 6 = 36 - 72 = -56 < 0.
Therefore, the roots of the given equation are not real.
2. Find the value of ‘p’, if the roots of the following quadratic equation are equal (p - 3)x\(^{2}\) + 6x + 9 = 0.
Solution:
For the equation (p - 3)x\(^{2}\) + 6x + 9 = 0;
a = p - 3, b = 6 and c = 9.
Since, the roots are equal
Therefore, b\(^{2}\) - 4ac = 0
⟹ (6)\(^{2}\) - 4(p - 3) × 9 = 0
⟹ 36 - 36p + 108 = 0
⟹ 144 - 36p = 0
⟹ -36p = - 144
⟹ p = \(\frac{-144}{-36}\)
⟹ p = 4
Therefore, the value of p = 4.
3. Without solving the equation 6x\(^{2}\) - 7x + 2 = 0, discuss the nature of its roots.
Solution:
Comparing 6x\(^{2}\) - 7x + 2 = 0 with ax\(^{2}\) + bx + c = 0 we have a = 6, b = -7, c = 2.
Therefore, discriminant = b\(^{2}\) – 4ac = (-7)\(^{2}\) - 4 ∙ 6 ∙ 2 = 49 - 48 = 1 > 0.
Therefore, the roots (solution) are real and unequal.
Note: Let a, b and c be rational numbers in the equation ax\(^{2}\) + bx + c = 0 and its discriminant b\(^{2}\) - 4ac > 0.
If b\(^{2}\) - 4ac is a perfect square of a rational number then \(\sqrt{b^{2} - 4ac}\) will be a rational number. So, the solutions x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) will be rational numbers. But if b\(^{2}\) – 4ac is not a perfect square then \(\sqrt{b^{2} - 4ac}\) will be an irrational numberand as a result the solutions x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) will be irrational numbers. In the above example we found that the discriminant b\(^{2}\) – 4ac = 1 > 0 and 1 is a perfect square (1)\(^{2}\). Also 6, -7 and 2 are rational numbers. So, the roots of 6x\(^{2}\) – 7x + 2 = 0 are rational and unequal numbers.
Quadratic Equation
Introduction to Quadratic Equation
Formation of Quadratic Equation in One Variable
Solving Quadratic Equations
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
Roots of a Quadratic Equation
Examine the Roots of a Quadratic Equation
Problems on Quadratic Equations
Quadratic Equations by Factoring
Word Problems Using Quadratic Formula
Examples on Quadratic Equations
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Quadratic Formula
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring
9th Grade Math
From Examine the Roots of a Quadratic Equation to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.