A quadratic equation has two roots with an unknown variable x. Let us learn about the roots of a quadratic equation with examples in this article. To find the roots of the quadratic function we set f(x) = 0 and then solve the quadratic equation. The roots of the equation can be equal real numbers or unequal real numbers or complex numbers.
Roots of a Quadratic Equation
The values of variables satisfying the quadratic equation are known as the roots of the equation. For every quadratic equation, there can be more than one solution. In order to find the quadratic equation, we have to use the standard form i.e, ax²+bx+c = 0. α and β are the unknown roots of the equation.
- One of the solutions of the quadratic equation is 0 and the other is -b/a in case if c = 0.
- Both the solutions are zero if b = c = 0.
- The roots of the equations are reciprocal to each other if a = c.
Discriminant
The discriminant is a part of the quadratic formula that depends upon the coefficient and properties of the roots of the equation. The discriminant will be zero only if the polynomial has double roots. It uncloses the nature of the roots of a quadratic equation.
The discriminant of the quadratic equation ax²+bx+c = 0 is,
D = b² – 4ac
Roots of Quadratic Equation Alpha Beta
α and β are the representation of roots of the quadratic equation.
The sum of the roots of the equation = α + β = -b/a
Product of the roots of the equation = αβ = c/a
The quadratic equation in the form of roots is x² – (α + β)x+ (αβ) = 0
The Nature of Roots of a Quadratic Equation
- D > 0, roots are real and distinct
- D = 0, roots are real and equal
- D < 0, roots are imaginary and unequal
- The quadratic equation will have real roots.
- The quadratic equation will have rational roots.
- Quadratic Equations will have integral roots.
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Roots of Quadratic Equation Examples
Example 1.
If α and β are the roots of the quadratic equation x² – 3x + 2 = 0. Find the sum of the roots and product of the roots.
Solution:
Given, x² – 3x + 2 = 0
a = 1, b = -3, c = 2
i. The sum of the roots = α + β = -b/a
α + β = -b/a = -(-3)/1 = 3
ii. Product of the roots of the equation = αβ = c/a
αβ = c/a = 2/1 = 2
Example 2.
If α and β are the roots of the quadratic equation 2x² – 6x + 10 = 0. Find the sum of the roots and product of the roots.
Solution:
Given the quadratic equation 2x² – 6x + 10 = 0
a = 2, b = -6, c = 10
We know that,
i. The sum of the roots = α + β = -b/a
α + β = -b/a = -(-6)/2 = 3
ii. Product of the roots of the equation = αβ = c/a
αβ = c/a = 10/2 = 5
Example 3.
Given that x² + (k – 5)x – k = 0 has real roots which differ by 4, determine,
i. the value of each root
ii. the value of k
Solution:
Let α be the smaller real root, then the other will be (α + 4).
The sum of the roots = α + (α + 4) = 2α + 4
The product of the roots is α(α + 4)
We know that,
i. The sum of the roots = α + β = -b/a
α + β = -b/a = – (k – 5)
ii. Product of the roots of the equation = αβ = c/a
αβ = c/a = -k
Sum of the roots
2α + 4 = -(k – 5)
2α + 4 = -k + 5
k = 1 – 2α
Product of the roots
α(α + 4) = -k
k = -α(α + 4)
-α² – 4α = 1 – 2α
α² + 2α + 1 = 0
(α + 1)(α + 1) = 0
α = -1
k = 1 – 2α
k = 1 – 2(-1)
k = 1 + 2
k = 3
Thus the roots of the equation are -1 and 3.
Example 4.
If the coefficient of x in the quadratic equation x² + bx + c =0 was taken as 17 in place of 13, its roots were found to be -3 and -14. Find the roots of the original quadratic equation.
Solution:
The product of zeros(c) = -3 × -14 = 42
Sume of the roots = -b/a = -13
x² – (Sum of Zeros)x + (Product of Zeros) = 0
x² – (-13)x + 42 = 0
x² + 13x + 42 = 0
x² + 7x + 6x + 42 = 0
x(x + 7) + 6(x + 7) = 0
(x + 6)(x + 7) = 0
x + 6 = 0
x = -6
x + 7 = 0
x = -7
Thus the roots of the equation is -6 and -7.
Example 5.
Form a quadratic equation with real coefficients when one of its roots is (2 – i).
Solution:
Since the complex roots always occur in pairs, so the other root is 2 + i.
Thus by obtaining the sum of the roots and product of the roots, we can form the essential quadratic equation.
The sum of the roots is
(2 + i) + (2 – i) = 4
The product of the root is (2 + i) × (2 – i)
= 2(2 – i) + i(2 – i)
4 – 2i + 2i – i²
4 – i²
4 – (-1) = 5
Hence, the equation is x² – (Sum of the roots)x + Product of the roots = 0
Thus the equation is x² – 4x + 5 = 0
FAQs on Roots of Quadratic Equations
1. Why do we find the roots of equations?
To find the roots of equations is to end the sets of equalities and also to understand the inequalities.
2. What is the root of the equation?
The root of the equation or solution of an equation is nothing but substituting the unknown variable in the equation.
3. How do you find the number of roots in an equation?
To find the number of roots in an equation we have to make use of the standard form of the equation ax²+bx+c = 0 and calculate the discriminant. If the discriminant D < 0 the quadratic had no real roots. If D = 0 then the quadratic has equal roots. If D > 0 then the equation has two distinct roots.