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[ended]Mathematics Tutoring Session #15 (Roots of a Quadratic Equation[Mon. 7th Mar 2016, 4:00 - 5:00pm])
Mathematics
Emmanuel Iwara(Tutor)
07-03-2016 13:22:00 +0000Good day all!
This session has ended for today
The nature of the roots of a quadratic equation is often tested by jamb and other exams.So let quickly take a look at following.
Discriminant (D) - This is the expression under the square root in the quadratic formula. The discriminant determines the nature of the roots of a quadratic equation. The word '(nature)' refers to the types of numbers the roots can be — namely real, rational, irrational or imaginary.
x = -b±√[(b2 - 4ac) / 2a]-------------------------(1)
D = [(b2 - 4ac) ]-------------------------(2)
(2) is called discriminant, which discriminates and analysis the nature of a root.
Nature of roots of a Quadratic Equation
When discriminant = b2 -4ac, then two cases arises
Case I.
When a, b, c are real numbers, a ≠0:
If = b2 -4 a c = 0, then roots are equal (and real).
If = b2 -4 a c > 0, then roots are real and unequal.
If = b2 -4 a c < 0, then roots are complex. It is easy to see that roots are a pair of complex conjugates.
Case II. When a, b, c are rational numbers, a ≠0:
If = b2 -4 a c = 0, then roots are rational and equal.
If = b2 -4 a c > 0, and is a perfect square of a rational number, then roots are rational and unequal.
If = b2 -4 a c > 0 but is not a square of rational number, then roots are irrational and unequal. They form a pair of irrational conjugates p +q, p - q where p, q Q, q> 0.
If =b2 - 4 a c <0, then roots are a pair of complex conjugates.
Illustrative Examples
Example 1.
Discuss the nature of the roots of the following equations:
(i) 4x2-12x + 9 = 0
(ii) 3x2 -10x + 3 = 0
(iii) 9x2 -2 = 0
(iv) x2 + x + 1 = 0
Solution (i) b = -12, a = 4, c = 9
= b2 -4 a c = (-12)2 -4 (4)(9) = 144 -144 = 0.
Hence the roots are rational and equal.
(ii) b =-10, a = 3, c = 3
= b2 -4 a c = (-10)² -4 (3)(3) = 100 -36 = 64.
Now = 64 > 0, and 64 is a perfect square of a rational number.
Hence the roots are rational and unequal.
(iii) b = 0, a = 9, c = -2
= b2 -4 a c = (0)2 -4 (9)(-2) = 72. Now = 72 > 0 but is not a perfect square of a rational number. Hence the roots are irrational and unequal.
(iv) b= 1, a = 1, c = 1
= b2 - 4 a c = (1)2 -4 (1)(1) = -3 < 0. Hence the roots are a pair of complex conjugates.
Example 2.
Find m so that the roots of the equation (4 + m)x2 + (m +1)x + 1 = 0 may be equal.
Solution
Using the formula b2 - 4 a c = 0 -------------------------------(*)
a = (4 + m), b = (m + 1), c = 1. into equation (*)
b2 = 4ac
(m + 1)2 = 4(4+m)(1)
[m2 + 2m + 1] = 16 + 4m
m2 + 2m -4m + 1 -16 = 0
m2 - 2m -15 = 0--------------------------(1)
using the formula method to solve for (m) in equation (1)
m = -3, 5.
Sum and Product of Roots
If the sum of a quadratic equation is given by β + α = -b/a---------------(1)
and the product as (β)(α) = c/a-----------------------------------(2)
Then the quadratic equation can be written as x2 -(β + α)x +(β)(α) = 0 ---------------------(3)
Example 3. Find the equation whose roots are (-3, 2)
Solution
Sum of roots = (-3 + 2) = -1 = β + α
Product of the roots = (-3)(2) = -6 = (α)(β)
Using equation (3), we have x2 -(-1)x + (-6) = 0
= x2 + x - 6 = 0
Now you may try to solve the following past jamb and other exams question;
Exercise 1. Find the nature of roots of the following equations.
(i) x2 + 9 = 0
(ii) x2 -22x + 1 = 0
2.For what value of k will the equation have equal root [x2 -2x + k -4] = 0
3.Find the equation whose roots are (-5, -4)
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Emmanuel Iwara(Tutor)
Hello class your tutor is around feel free to ask your questions if any i will try the best i can to answer them.
0 07-03-2016 15:05:00 +0000
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Emmanuel Iwara(Tutor)
@Rhoda Otaru the sum of root = (alpha + beta) = (-5 + -4) = -9, product root = (-5 x-4) = 20 Then the quadratic equation can be written as x^2 -(β + α)x +(β)(α) = 0 x^2 - (-9)x + 20 = 0, hence x^2 + 9x + 20 = 0 ,the required equation
0 07-03-2016 16:20:00 +0000
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