FINDING UNKNOWN THROUGH SUM AND PRODUCT OF ROOTS

If α, β, are the roots of the quadratic equation, then the form of the quadratic equation as

x2 – (α + β)x + αβ = 0

Where,

α + β = sum of roots

αβ = product of roots

Problem 1 :

Find k if the difference between the roots of the quadratic equation

x2 – 4x + k = 0 is 2.

Solution :

x2 – 4x + k = 0

a = 1, b = -4, c = k

α – β = 2 ----- (1)

α + β = -b/a

α + β = -(-4)/1

α + β = 4 ----- (2)

α – β = 2

α + β = 4

(1) + (2)

2α = 6

α = 6/2

α = 3

Products of the roots = αβ = c/a

 β = k/1

β = k/3

α – β = 2

3 – (k/3) = 2

-k/3 = 2 – 3

-k/3 = -1

k = 3

So, the value of k is 3.

Problem 2 :

Find the value of p such that the difference of the roots of the equation

x2 – px + 8 = 0 is 2.

Solution :

x2 – px + 8 = 0

a = 1, b = -p, c = 8

Difference between the roots = α – β = 2 ----- (1)

Sum of the roots = α + β = -b/a

α + β = -(-p)/1

α + β = p ----- (2)

Product of roots = α β = 8

(1) + (2)

2α = p + 2

α = (p + 2)/2

Applying the value of α = (p + 2)/2 in α β = 8

β = 8 ⋅ 2/(p + 2) 

β = 16/(p + 2)

Applying the values of α and β in (1), we get

(p + 2)/2 - 16/(p + 2) = 2

p +22 - 16p + 2 = 2(p +2)2 - 322(p+2)= 2p2+4p+4-32=4p+8p2-28=8p2= 36p = ±6

So, the value of p is -6 or 6.

Problem 3 :

Find the value of k such that the difference of the roots of the equation

2kx2 – 20x + 21 = 0 is 2.

Solution :

2kx2 – 20x + 21 = 0

a = 2k, b = -20, c = 21

Difference between the roots = α – β = 2 ----- (1)

α = 2 + β

α + β = -b/a

α + β = -(-20)/2k

α + β = 10/k ----- (2)

α β = c/a

α β = 21/2k ---(3)

Applying the value of α in (2) and (3), we get

2 + β + β = 10/k

2 + 2β = 10/k

2β = (10/k) - 2

β = (5/k) - 1

(2 + β) β = 21/2k

Applying β

[2 + (5/k) - 1][(5/k) - 1] = 21/2k

((5/k) - 1)((5/k) + 1) = 21/2k

25/k2 - 1 = 21/2k

25k2 - 1 = 212k25k2 = 212k+125k2 = 21 + 2k2k50 = k(21+2k)50 = 21k+2k22k2 + 21k -50 = 0(2k + 25) (k - 2)= 0k = -252 and k = 2

Problem 4 :

Find k so that one root of the equation 2x2 – 16x + k = 0 is twice the other. (Hint : One root = α, Other root = 2α)

Solution :

Given, 2x2 – 16x + k = 0

a = 2, b = -16, c = k

α = α, β = 2α

α + β = -b/a

α + 2α = -(-16)/2 = 16/2

3α = 8

α = 8/3

αβ = c/a

α(2α) = k/2

2α2 = k/2

α2 = k/4

By applying the value of α, we get

(8/3)2 = k/4

64/9 = k/4

k = (64 ⋅ 4)/9

k = 256/9

So, the value of k is 256/9.

Problem 5 :

Find k so that one root of the equation

k(x – 1)2 = 5x – 7

is twice the other.

Solution :

k(x – 1)2 = 5x – 7

k(x2 + 1 – 2x) – 5x + 7 = 0

kx2 + k – 2xk – 5x + 7= 0

x2(k) - x(2k + 5) + k + 7 = 0

a = k, b = 2k + 5, c = k + 7

α = α, β = 2α

α + β = -b/a

 α + 2α =( 2k + 5)/k

3α = (2k + 5)/k

α = (2k + 5)/3k ---(1)

αβ = c/a

α(2α) = (k + 7)/k

2α2 = (k + 7)/k

α2 = (k + 7)/2k ---(2)

By applying α in (2), we get

2k+53k2= k+72k(2k+5)29k2= k+72k(2k+5)29k= k+722(2k+5)2 = 9k(k + 7)24k2+20k+25 = 9k2+63k 8k2+40k+50 = 9k2+63k k2+23k-50=0(k + 25)(k - 2) = 0

Equating each factor to zero, we get

k = -25 and k = 2

Problem 6 :

Find k so that the roots of the quadratic equation

2x2 + 3x + k = 0

are equal. 

Solution :

Given, 2x2 + 3x + k = 0

a = 2, b = 3, c= k

α = α, β = α

α + β = -b/a

α + α = -3/2

2α = -3/2

α = -3/2 × 1/2

α = -3/4

αβ = c/a

α × α = k/2

-3/4 × -3/4 = k/2

9/16 = k/2

9/16 × 2 = k

9/8 = k

So, the value of k is 9/8.

Problem 7 :

If 1 – i and 1 + i are the roots of the equation

x2 + ax + b = 0

where a, b ∈  r, then find the values of a and b.

Solution :

Given, x2 + ax + b = 0

a = 1, b = a, c = b

α = 1 - i, β = 1 + i

α + β = -b/a

1 – i + 1 + i = -a

2 = -a

a = -2

αβ = c/a

(1 – i) × (1 + i) = b

1 – i2 = b

1 + 1 = b

2 = b

So, the values of a and b is -2 and 2 respectively.

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