# FIND THE OTHER ROOT OF A QUADRATIC EQUATION IF ONE ROOT IS GIVEN

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

we can figure out the other root.

Problem 1 :

If a root of the equation

x2 – 6x + k = 0 is 4

find the second root and the missing value.

Solution :

x2 – 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.

x2 – 6x + 8 = 0

Using factorization.

x2 – 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

x2 + kx + 18 = 0

with one root of 6, find the second root and the missing value.

Solution :

x2 + 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.

x2 - 9x + 18 = 0

Using factorization.

x2 - 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 :

x2 - x + k = 0, r1 = -4

Solution :

x2 - x + k = 0

Let x = r1

r12 – r1 + k = 0 ----- (1)

One root r1 = -4.

Substituting r1 = -4 in equation (1)

(-4)2 – (-4) + k = 0

16 + 4 + k = 0

20 + k = 0

k = -20

k = -20 substitute in given equation.

x2 - x - 20 = 0

Using factorization.

x2 – 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 :

2x2 + bx –15 = 0, r1 = 3

Solution :

2x2 + bx –15 = 0

Let r1 and r2 be the two roots

Here r1 = 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.

2x2 - x –15 = 0

Using factorization.

2x2 – 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 :

3x2 - x + k = 0, r1 = -5/3

Solution :

3x2 - x + k = 0

Let x = r1

One root r1 = -5/3.

Substituting r1 = -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.

3x2 - x - 10 = 0

Using factorization.

3x2 – 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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