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A quadratic function is a polynomial function in one variable having degree 2. When the function is equated to zero, it is called a quadratic equation. It is represented in the standard form f(x)= ax2 + bx + c= 0, where a, b and c are real numbers and a ≠ 0. The coefficient a is called the lead or quadratic coefficient, b the linear coefficient, c the constant or the free term and x the variable.
Application of quadratic equations
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The quadratic equation appears in a lot of real-world problems like determining:
- Dimensions and areas of land, boxes etc. subjected to conditions,
- Cost of product and profit (where the optimum price of a product that will yield the most profit at a given demand is represented by the vertex of the parabola of a quadratic equation)
- The trajectory of thrown objects like a ball or a bomb, as thrown objects follow a parabolic path and quadratic equations can be used to determine the position of the object along the path.
- Speed in cruises. Kayakers use quadratic equations to calculate the speed while rowing upstream and downstream.
- Values of resistors connected in a parallel circuit in electronics.
The solution to the above problems can be found by resolving for the variable x using algebraic or graphical methods. The algebraic methods are:
- Factorization method
- Quadratic formula method
- Completing the square method
The values of the variable x which satisfy the conditions of the equation is called a solution or root of the quadratic equation. The number of roots of a polynomial equation is always equal to its degree and hence, a quadratic equation (with degree 2) will have 2 roots.
Graphically, the quadratic equation is represented by a parabola. If a > 0, then the parabola is upward-pointing. If a < 0, the parabola is downward pointing. The maximum or the minimum point on the parabola, where it turns around, is called the vertex. The solution or root will be any value of x where the parabola intersects the x-axis. This is the point where the value on the y-axis will be 0.
The given formula helps to obtain the roots of a quadratic equation:
\(x=\frac{(-b\pm \sqrt{(b^{2}-4ac)})}{2a}\)Where D is called the Discriminant
is given as:
D = b2 - 4ac
The video below explains this:
Quadratic Equations Detailed Video Explanation:
Nature of roots of a quadratic equation
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The nature of the roots of a quadratic equation can be any of the following:
- Real and distinct
- Real and equal
- Complex and distinct
The nature of the roots can be determined by calculating the Discriminant as per the above formula.
When,
D > 0, the roots will be real and distinct with the parabola intersecting the x-axis at two separate points. If D is a perfect square, the roots are rational and if D is not a perfect square the roots are irrational.
D = 0, the roots will be real and equal with the parabola intersecting the x-axis only at one point. This happens when the vertex of the parabola intersects the x-axis.
D < 0, the roots will be complex and distinct and the parabola does not intersect the x-axis at all.
Examples:
Let’s go through some examples to determine the nature of the roots using the value of the Discriminant and calculate the roots for the quadratic equations.
- Example 1: f(x) = 6x2 - x - 15 = 0
Calculating the Discriminant, D = b2 - 4ac
= -12 – (4 x 6 x -15)
= 1 – (- 360)
D = 361
Since, D > 0, the equation has real and distinct roots.
The roots will be, \(x=\frac{(-b+\sqrt{D})}{2a}=\frac{[-(-1)+\sqrt{361}]}{(2\times 6)}=\frac{5}{3}\)
\(x=\frac{(-b-\sqrt{D})}{2a}=\frac{[-(-1)-\sqrt{361}]}{(2\times 6)}=\frac{-3}{2}\)
x = \(\frac{5}{3}\) and \(\frac{-3}{2}\)
Here, the roots of the equation are real numbers and distinct.
- Example 2: f(x) = x2 + 2x + 1 = 0
Calculating the Discriminant, D = b2 - 4ac
= 22 – (4 x 1 x 1)
= 4 – 4
D = 0
Since, D = 0, the equation has real and equal roots.
The roots will be, \(x=\frac{(-b+\sqrt{D})}{2a}=\frac{[(-2)+\sqrt{0}]}{(2\times 1)}=\frac{-2}{2}=-1\)
\(x=\frac{(-b-\sqrt{D})}{2a}=\frac{[(-2)-\sqrt{0}]}{(2\times 1)}=\frac{-2}{2}=-1\)
x = -1 and -1
Here, the roots of the equation are real and equal numbers.
- Example 3: f(x) = 3x2 - 4x + 2 = 0
Calculating the Discriminant, D = b2 - 4ac
= (-4)2 – (4 x 3 x 2)
= 16 – 24
D = -8
Since, D < 0, the equation has a complex root.
The roots will be, \(x=\frac{(-b+\sqrt{D})}{2a}=\frac{[-(-4)+\sqrt{-8}]}{(2\times 3)}\)
\(\frac{[4+2\sqrt{-2}]}{(6)}=\frac{[2+\sqrt{2}i]}{(3)}\)
\(x=\frac{(-b-\sqrt{D})}{2a}=\frac{[-(-4)-\sqrt{-8}]}{(2\times 3)}\)
\(\frac{[4-2\sqrt{-2}]}{(6)}=\frac{[2-\sqrt{2}i]}{3}\)
x = \(\frac{[2+\sqrt{2}i]}{(3)}\) and \(\frac{[2-\sqrt{2}i]}{3}\)
Here, the roots of the equation are complex numbers.
Sample Questions
Ques: What are the roots of a quadratic equation and explain its nature?(2 marks)
Ans. x2 - x - 2 = 0
When we compare with ax2 + bx + c = 0, we obtain,
a = 1, b = -1, c = -2
Hence, D = b2- 4ac = 1 - 4(1)(-2) = 1+ 8 = 9
D>0
Therefore we can say that the equation has two prominent real roots.
Ques: What is the Discriminant for the quadratic equation x2 - 3x + 2 = 0?(2 marks)
Ans. The discriminant’s value of x2 - 3x + 2 = 0 is
D = b2 - 4ac
A = 1, b = -3, c = 2
Therefore D = (-3)² - 4 \(\times\) 1 \(\times \) 2
= 9 - 8
= 1
Ques: Find the Discriminant of the quadratic equation 3x2 – 2x + 1/3 = 0 and find the nature of its roots?(2 marks)
Ans. The discriminant of a quadratic equation determines its roots = D = b2 - 4ac
When we compare 3x2 - 2x + \(\frac{1}{3}\) = 0 with ax² + bx + c = 0 we obtain a = 3, b = -2, c = \(\frac{1}{3}\)
Therefore D = b2 - 4ac
= (-2)2 - 4 * 3 * \(\frac{1}{3}\)
= 4- 4
= 0
Hence we can say the roots are equal and real
The roots are as follows: \(\frac{(-b+(-b)^{2}-4ac)}{2a}\)
= \(\frac{-(-2)+(-2)^{2}-4\times 3\times \frac{1}{3})}{2\times 3}\)
= \(\frac{2}{{2\times 3}}\)
= \(\frac{1}{3}\)
Therefore roots are as follows: \(\frac{1}{3}\), \(\frac{1}{3}\)
Ques: What should be the value of K for the quadratic equation kx2 + 2kx + 6 = 0 to have real roots(2 marks)
Ans. When the quadratic equation has equal roots, then b2 - 4ac = 0
We know kx2 - 2kx + 6 = 0
We have a = k, b = -2k, c = 6
Now, (-2k)2 - 4 \(\times\) k \(\times\) 6 = 0
4k2 - 24k = 0
4k (k - 6) = 0
k = 0 or k = 6
Ques: What should be the value of K for the quadratic equation 4x2 + kx + 3= 0 to have equal roots?(2 marks)
Ans. We have, 4x2 + kx + 3= 0
When we put it like ax2 + bx + c = 0
Here a = 4, b = k, c = 3
When we want the quadratic equation we have to have real roots D = b2 - 4ac ≥ 0
D = (k)2 - 4(4)(3) ≥ 0
k2 - 48 ≥ 0
k2 ≥ 48
k ≥ 4√3and k ≤ -4√3 {After we take the sq. roots on both the sides}
We can represent the value of k as = [(∞, \(4\sqrt{3}\)) U ( -\(4\sqrt{3}\), -∞)]
Previous Year Questions
Ques: For what values of k, the roots of the equation x2 + 4x + k = 0 are real? (2019)
Ans.
Given x2 + 4x + k = 0
When we compare the equation with ax2 + bx + c = 0 we get,
A = 1, b = 4, c = k
For real roots D ≥ 0
Or b2 - 4ac ≥ 0
Or 16 - 4k ≥ 0
Or k ≤ 4
When k ≤ 4, the equation x2 + 4x + k will have real roots.
Ques: Find the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other. (2019)
Ans. Given 3x2 – 10x + k = 0
When we compare the equation with ax2 + bx + c = 0 we get,
a = 3, b = -10, c = k
The product of the roots = \(\frac{k}{{3}}\)
\(a \times \frac{1}{a}=\frac{k}{3}\)
1 = \(\frac{k}{{3}}\)
Or k = 3
Ques: Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0. (outside Delhi 2019)
Ans. We have 2x2 – 4x + 3 = 0
When we compare the equation with ax² + bx + c = 0 we get here,
a = 2, b = -4, c = 3
D = b2 - 4ac = (-4)² - 4 \(\times\) (2)(3) = 16 - 24 = -8 < 0
Therefore the roots will be imaginary is shown by D < 0
Ques: If x = 3 is one root of the quadratic equation x2– 2kx – 6 = 0, then find the value of k. (2018)
Ans. Given x2– 2kx – 6 = 0
x = 3 is the root of the given quadratic equation, then,
(3)2 - 2k(3) - 6 = 0
9 - 6k - 6 = 0
3 - 6k = 0
3 = 6k
k = \(\frac{3}{{6}}\) = \(\frac{1}{{2}}\)
k = \(\frac{1}{{2}}\)
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