For any quadratic equation, α and β be two roots. The given root can be considered as a value of the variable in which they have defined.
Once we create the quadratic equation completely, by using
(i) Factorization
(ii) Quadratic formula
we can figure out the other root.
Problem 1 :
If a root of the equation
x^{2} – 6x + k = 0 is 4
find the second root and the missing value.
Solution :
x^{2} – 6x + k = 0
One root of 4.
Substituting x = 4 in the given equation.
(4)^{2} – 6(4) + k = 0
16 – 24 + k = 0
-8 + k = 0
k = 8
k = 8 substitute in given equation.
x^{2} – 6x + 8 = 0
Using factorization.
x^{2} – 4x – 2x + 8 = 0
x(x – 4) – 2(x – 4) = 0
x – 2 = 0 and x – 4 = 0
x = 2 and x = 4
So, the second root is x = 2.
Problem 2 :
Given the equation
x^{2} + kx + 18 = 0
with one root of 6, find the second root and the missing value.
Solution :
x^{2} + kx + 18 = 0
One root of 6.
Substituting x = 6 in given equation.
(6)^{2} + k(6) + 18 = 0
36 + 6k + 18 = 0
54 + 6k = 0
6k = -54
k = -54/6
k = -9
k = -9 substitute in given equation.
x^{2} - 9x + 18 = 0
Using factorization.
x^{2} - 6x - 3x + 18 = 0
x(x – 6) – 3(x – 6) = 0
(x – 3) (x – 6) = 0
x – 3 = 0 and x – 6 = 0
x = 3 and x = 6
So, the second root is x = 3.
For these equations, one root is given. Find the second root and the missing value.
Problem 3 :
x^{2} - x + k = 0, r_{1} = -4
Solution :
x^{2} - x + k = 0
Let x = r_{1}
r_{1}^{2} – r_{1} + k = 0 ----- (1)
One root r_{1} = -4.
Substituting r_{1} = -4 in equation (1)
(-4)^{2} – (-4) + k = 0
16 + 4 + k = 0
20 + k = 0
k = -20
k = -20 substitute in given equation.
x^{2} - x - 20 = 0
Using factorization.
x^{2} – 5x + 4x – 20 = 0
x(x – 5) + 4(x – 5) = 0
(x + 4) (x – 5) = 0
x + 4 = 0 and x – 5 = 0
x = -4 and x = 5
So, the second root is x = 5.
Problem 4 :
2x^{2} + bx –15 = 0, r_{1} = 3
Solution :
2x^{2} + bx –15 = 0
Let r_{1} and r_{2 }be the two roots
Here r_{1} = 3
By applying one root, we get
2(3)^{2} + b(3) –15 = 0
18 + 3b – 15 = 0
3 + 3b = 0
3b = -3
b = -1
b = -1 substitute in given equation.
2x^{2} - x –15 = 0
Using factorization.
2x^{2} – 6x + 5x -15 = 0
2x(x – 3) + 5(x – 3) = 0
(2x + 5) (x – 3) = 0
2x + 5 = 0 and x – 3 = 0
2x = -5 and x = 3
x = -5/2
So, the second root is x = -5/2.
Problem 5 :
3x^{2} - x + k = 0, r_{1} = -5/3
Solution :
3x^{2} - x + k = 0
Let x = r_{1}
One root r_{1}
= -5/3.
Substituting r_{1} = -5/3 in equation
3(-5/3)^{2} – (-5/3) + k = 0
3(25/9) + 5/3 + k = 0
75/9 + 5/3 + k = 0
75/9 + 5/3 × (3/3) + k = 0
75/9 + 15/9 + k = 0
90/9 + k = 0
10 + k = 0
k = -10
k = -10 substitute in given equation.
3x^{2} - x - 10 = 0
Using factorization.
3x^{2} – 6x + 5x -10 = 0
3x(x – 2) + 5(x – 2) = 0
(3x + 5) (x – 2) = 0
3x + 5 = 0 and x -2 = 0
3x = -5 x =2
x = -5/3
So, the second root is x = 2.
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