We use sum and product of roots to find the unknown involving in the quadratic equation.
A general form of a quadratic equation ax^{2} + bx + c = 0
To find the sum and product of the roots of the quadratic equation,
Sum of the roots = -b/a
Products of the roots = c/a
If α, β, are the roots of the quadratic equation, then the form of the quadratic equation as
x^{2} – (α + β)x + αβ = 0
Where,
α + β = sum of roots
αβ = product of roots
Use the sum and product of roots formulas to answer the questions below :
Problem 1 :
The roots of the equation x^{2} – kx + k – 1 = 0 are α and 2α. Find the value(s) of k.
Solution :
x^{2} – kx + k – 1 = 0
a = 1, b = -k, c = k - 1
α = α, β = 2α
α + β = -b/a α + 2α = -(-k)/1 = k 3α = k α = k/3 ---(1) |
αβ = c/a α × 2α = (k – 1)/1 2α^{2} = k – 1 ---(2) |
Applying (1) in (2), we get
2(k/3)^{2} = k – 1
2k^{2}/9 = k – 1
2k^{2} = 9(k – 1)
2k^{2} = 9k – 9
2k^{2} – 9k + 9 = 0
(k – 3) (2k – 3) = 0
k = 3, 2k = 3
k = 3/2
So, the value(s) of k is 3 and 3/2.
Problem 2 :
The roots of the quadratic equation
x^{2} + 6x + c are k and k - 1
Find the value of c.
Solution :
x^{2} + 6x + c
a = 1, b = 6, c = c
α = k, β = k - 1
α + β = -b/a k + (k – 1) = -6/1 = -6 2k – 1 = -6 2k = -6 + 1 2k = -5 k = -5/2 ---(1) |
αβ = c/a k × (k – 1) = c/1 = c k^{2} – k = c ---(2) |
Applying (1) in (2), we get
(-5/2)^{2} – (-5/2) = c
25/4 + 5/2 = c
25/4 + 5/2 × (2/2) = c
25/4 +10/4 = c
35/4 = c
So, the value of c is 35/4.
Problem 3 :
The roots of the quadratic equation
2x^{2} - 9x + k
are m/2 and m - 3. Find the value of k.
Solution :
2x^{2} - 9x + k
a = 2, b = -9, c = k
α = m/2, β = m - 3
α + β = -b/a = -(-9)/2 m/2 + (m – 3) = 9/2 m + 2m – 6 = (9/2) × 2 3m – 6 = 9 3m = 9 + 6 m = 15/3 m = 5 |
αβ = c/a m/2 × (m – 3) = k/2 5/2 × (5 – 3) = k/2 5/2 × 2 = k/2 5 = k/2 10 = k |
So, the value of k is 10.
Problem 4 :
Find the values of m for which one root of the equation
4x^{2} + 5 = mx
is three times the other root.
Solution :
4x^{2} + 5 = mx
4x^{2} - mx + 5 = 0
a = 4, b = -m, c = 5
Let, the roots are α and 3α.
α = α, β = 3α
α + β = -b/a α + 3α = -(-m)/4 4α = m/4 α = m/4 × 1/4 α = m/16 ---(1) |
αβ = c/a α × 3α = 5/4 3α^{2} = 5/4 α^{2} = 5/4 × 1/3 α^{2} = 5/12 ---(2) |
By applying (1) in (2), we get
(m/16)^{2} = (5/12)
m^{2}/256 = (5/12)
m^{2} = 5/12 × 256
m^{2} = 320/3
m =√(320/3)
So, the value of m is 8√(5/3).
Problem 5 :
One root of the equation 3x^{2} - 4x + m = 0 is double the other. Find the roots, and the value of m.
Solution :
3x^{2} - 4x + m = 0
a = 3, b = -4, c = 1
α = 2β, β = β
α + β = -b/a 2β + β = -(-4)/3 3β = 4/3 β = 4/3 × 1/3 β = 4/9 ---(1) |
αβ = c/a 2β × β = m/3 2β^{2} = m/3 β^{2} = m/3 × 1/2 β^{2} = m/6 ---(2) |
By applying (1) in (2), we get
(4/9)^{2} = m/6
16/81 = m/6
16/81 × 6 = m
96/81 = m
32/27 = m
So, the value of m is 32/27.
Problem 6 :
The roots of the equation
4x^{2} - kx + 35 = 0
differ by one. Find the value of k.
Solution :
4x^{2} - kx + 35 = 0
a = 4, b = -k, c = 35
α + β = -b/a = -(-k)/4 = k/4 |
αβ = c/a = 35/4 |
√((α + β)^{2} - 4αβ) = 1
√((k/4)^{2} – 4(35/4) = 1
√((k^{2}/16) - 35) = 1
Taking square on both sides.
((k^{2}/16) - 35) = 1
k^{2}/16 = 1 + 35
k^{2}/16 = 36
k^{2} = 36 × 16
k^{2 }= 576
k = √576
k = 24
So, the value of k is 24.
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