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

(ii) Quadratic formula

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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