# FINDING UNKNOWN IN A QUADRATIC EQUATION WHEN ROOTS ARE GIVEN

We use sum and product of roots to find the unknown involving in the quadratic equation.

A general form of a quadratic equation ax2 + 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

x2 – (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 x2 – kx + k – 1 = 0 are α and 2α. Find the value(s) of k.

Solution :

x2 – kx + k – 1 = 0

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

α = α, β = 2α

 α + β = -b/a α + 2α = -(-k)/1 = k3α = kα = k/3 ---(1) αβ = c/aα × 2α = (k – 1)/1 2α2 = k – 1 ---(2)

Applying (1) in (2), we get

2(k/3)2 = k – 1

2k2/9 = k – 1

2k2 = 9(k – 1)

2k2 = 9k – 9

2k2 – 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

x2 + 6x + c are k and k - 1

Find the value of c.

Solution :

x2 + 6x + c

a = 1, b = 6, c = c

α = k, β = k - 1

 α + β = -b/ak + (k – 1) = -6/1 = -62k – 1 = -62k = -6 + 12k = -5k = -5/2 ---(1) αβ = c/ak × (k – 1) = c/1 = ck2 – 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

2x2 - 9x + k

are m/2 and m - 3. Find the value of k.

Solution :

2x2 - 9x + k

a = 2, b = -9, c = k

α = m/2, β = m - 3

 α + β = -b/a= -(-9)/2m/2 + (m – 3) = 9/2m + 2m – 6 = (9/2) × 23m – 6 = 93m = 9 + 6m = 15/3m = 5 αβ = c/am/2 × (m – 3) = k/25/2 × (5 – 3) = k/25/2 × 2 = k/25 = k/210 = k

So, the value of k is 10.

Problem 4 :

Find the values of m for which one root of the equation

4x2 + 5 = mx

is three times the other root.

Solution :

4x2 + 5 = mx

4x2 - mx + 5 = 0

a = 4, b = -m, c = 5

Let, the roots are α and 3α.

α = α, β = 3α

 α + β = -b/aα + 3α = -(-m)/44α = m/4α = m/4 × 1/4α = m/16 ---(1) αβ = c/aα × 3α = 5/43α2 = 5/4α2 = 5/4 × 1/3α2 = 5/12 ---(2)

By applying (1) in (2), we get

(m/16)2 = (5/12)

m2/256 = (5/12)

m2 = 5/12 × 256

m2 = 320/3

m =√(320/3)

So, the value of m is 8√(5/3).

Problem 5 :

One root of the equation 3x2 - 4x + m = 0 is double the other. Find the roots, and the value of m.

Solution :

3x2 - 4x + m = 0

a = 3, b = -4, c = 1

α = 2β, β = β

 α + β = -b/a2β + β = -(-4)/33β = 4/3β = 4/3 × 1/3β = 4/9  ---(1) αβ = c/a2β × β = m/32β2 = m/3β2 = m/3 × 1/2β2 = 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

4x2 - kx + 35 = 0

differ by one. Find the value of k.

Solution :

4x2 - 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

√((k2/16) - 35) = 1

Taking square on both sides.

((k2/16) - 35) = 1

k2/16 = 1 + 35

k2/16 = 36

k2 = 36 × 16

k2 = 576

k = √576

k = 24

So, the value of k is 24.

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