Basic results related to a quadratic equation:
If p + √q is an irrational root of the quadratic relation, then p - √q is also a root of the quadratic equation given that all the coefficients are rational.
Problem: If α, β be the roots of the relation x2 + 2ax + b = 0, form a quadratic with rational with rational coefficients one of whose roots is a + b +
.
Key concept: Since the coefficients of the provided equation are rationals and one provided root is irrational, therefore its second root will also be irrational and conjugate of the provided root.
Solution: Given α + β = - 2a and αβ = b
We take that roots of the required equation can be α + β +
and α + β - .
therefore, the sum of the roots of the needed equation is
= 2(α + β) = - 4a, and the multiplication of the roots of the needed equation is= (α + β)2 - (α2 + β2) = 2αβ = 2b
therefore, the needed equation is
x2 + 4ax + 2b = 0 .
- The quadratic equation obtain rational roots if D is a perfect square and a, b, c are rational.
- If a, b, c are odd numbers, then the roots of quadratic equation may not be rational.
- If a = 1 and b, c are natural number and the roots of the quadratic equation are rational, then the roots have to be integers.
Problem: Calculate the total number of values of a so that x2 - x- a = 0 has integral roots , where a∈N, 6 ≤ a ≤ 100.
Key concept: Since each coefficients of the provided equation are integers, therefore for roots to be integer discriminant have to be a perfect square.
Solution: Now D = 1 + 4a = which is an odd integer.
therefore it will be in the form of D = (2λ + 1)2
=> 1 + 4a = 1 + 4λ4 + 4λ
=> a = λ(l + 1), therefore a could be in the form of multiplication of two consecutive integers. Since a ∈[6, 100], hence
a = 6, 12, 20, 30, 42, 56, 72, 90
So 'a' may contain 8 different values.
- A quadratic equation, whose roots are a and b may be defined as (x - a) (x - b) = 0. i.e., ax2 + bx + c ≡ a(x - a) (x - b).
Problem: Suppose a, b be the roots of the equation (x - a) (x - β) = c , c ≠ 0. Then calculate the roots of the equation (x - a)(x - b) + c = 0 in parts of a and b.
Solution: Since the roots of the equation(x - a) (x - b) - c=0 are a, b therefore that equation may be written as (x-a) (x-b) - c= (x-a)(x-b)
=> (x-a)(x-b) +c ≡ (x-a) (x-b).
Obviously roots of ( x-a ) ( x-b) + c = 0 are a and b.
Illustration: Calculate the equation whose roots are negative of the roots of the equation x3 -5x2 -7x -3=0.
Key concept: To get an equation whose roots are negative of the roots of a provided equation substitute x by -x.
Solution: (-x)3 -5 (-x)2 -7(-x) -3 = 0
or -x3 -5x2 + 7x -3 = 0 or x3 + 5x2 -7x + 3 = 0
If a + b + c = 0, then 1 is a root of ax2 + bx + c = 0.
If a + b + c = 0 and a, b, c are rational, then roots of equation ax2 + bx + c = 0 are rational.
Illustration: If the roots of (a - b)x2 + (b - c)x + (c - a) = 0 are same and real, then prove that 2a = b + c.
Key concept: Since the roots of the shown equation are same and real so discriminant has to be equal to zero.
Solution 1: Since roots of shown equation are same and real. therefore D = 0
=> (b - c)2 - 4(a - b)(c - a) = 0.
=> (2a - b - c)2 = 0 Þ 2a = b + c.
Key concept: Since addition of all the coefficients in the provided equation is zero therefore one will be one root of the provided equation. Now show that both the roots are same therefore both the roots may be one and their multiplication can also be same to one.
Illustration: If a, b, c, d are four non-zero real numbers so that (d + a - b)2 + (d + b - c)2 = 0 and roots of the relation a(b - c)x2 +b(c - a)x + c(a - b) = 0 are same and real, then defines that a=b=c.
Key concept: In the first provided equation we can use the idea that if
. And in the second equation since sum of all the coefficients is zero and provided both the roots are same product their roots will be same to one.
Solution: From the first equation(d + a - b)2 + (d + b - c)2 = 0
=> d + a - b=0 and d + b - c => 2b = a + c..............(1)
And from the second provided equation product of the roots is same to one.=>b = 2ac/a+c...(2)
From (1) and (2) 4ac/a+c = a + c Þ(a - c)2 = 0 => a = c
therefore a = b = c
Quadratic expression ax2 + bx + c is a complete square (i.e. square of a polynomial of degree one) if b2 - 4ac = 0.
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