Math Roots Explorer ✨

Discover the magic of quadratic equations and their roots!

Key Formulas to Remember:

For any quadratic equation x² - (α+β)x + αβ = 0:

📝 Problem 1: Express in terms of α+β and αβ
(i) α/(3β) + β/(3α)

Let's simplify the expression:

α/(3β) + β/(3α) = (α² + β²)/(3αβ)

We know that α² + β² = (α+β)² - 2αβ

So the expression becomes: [(α+β)² - 2αβ]/(3αβ)

(ii) 1/(α²β) + 1/(β²α)

Let's simplify the expression:

1/(α²β) + 1/(β²α) = (β + α)/(α²β²)

We can write this as: (α+β)/(αβ)²

(iii) (3α - 1)(3β - 1)

Let's expand the expression:

(3α - 1)(3β - 1) = 9αβ - 3α - 3β + 1

We can rewrite this as: 9αβ - 3(α+β) + 1

(iv) (α+3)/β + (β+3)/α

Let's simplify the expression:

(α+3)/β + (β+3)/α = (α² + 3α + β² + 3β)/(αβ)

We know that α² + β² = (α+β)² - 2αβ

So the expression becomes: [(α+β)² - 2αβ + 3(α+β)]/(αβ)

🔍 Problem 2: Roots of 2x² - 7x + 5 = 0

First, identify sum and product of roots:

α + β = -(-7)/2 = 7/2

αβ = 5/2

(i) 1/α + 1/β

We know that:

1/α + 1/β = (α + β)/(αβ)

Substituting the values: (7/2)/(5/2) = 7/5

(ii) α/β + β/α

We can write:

α/β + β/α = (α² + β²)/(αβ) = [(α+β)² - 2αβ]/(αβ)

Substituting the values: [(7/2)² - 2×(5/2)]/(5/2) = [49/4 - 5]/(5/2) = (29/4)/(5/2) = 29/10

(iii) (α+2)/(β+2) + (β+2)/(α+2)

Let's simplify the expression:

(α+2)/(β+2) + (β+2)/(α+2) = [(α+2)² + (β+2)²]/[(α+2)(β+2)]

Numerator: (α² + 4α + 4 + β² + 4β + 4) = (α²+β²) + 4(α+β) + 8

Denominator: αβ + 2(α+β) + 4

Substituting values:

Numerator: [(α+β)² - 2αβ] + 4(α+β) + 8 = (49/4 - 5) + 14 + 8 = 29/4 + 22 = 117/4

Denominator: 5/2 + 7 + 4 = 5/2 + 11 = 27/2

Final expression: (117/4)/(27/2) = (117×2)/(4×27) = 234/108 = 13/6

🌱 Problem 3: Roots of x² + 6x - 4 = 0

First, identify sum and product of roots:

α + β = -6

αβ = -4

(i) Equation with roots α² and β²

For new equation with roots α² and β²:

Sum: α² + β² = (α+β)² - 2αβ = (-6)² - 2(-4) = 36 + 8 = 44

Product: α² × β² = (αβ)² = (-4)² = 16

Equation: x² - (sum)x + product = 0

So: x² - 44x + 16 = 0

(ii) Equation with roots 2/α and 2/β

For new equation with roots 2/α and 2/β:

Sum: 2/α + 2/β = 2(α+β)/(αβ) = 2(-6)/(-4) = 3

Product: (2/α)(2/β) = 4/(αβ) = 4/(-4) = -1

Equation: x² - (sum)x + product = 0

So: x² - 3x - 1 = 0

(iii) Equation with roots α²β and β²α

First, note that α²β + β²α = αβ(α+β) and α²β × β²α = (αβ)³

Sum: αβ(α+β) = (-4)(-6) = 24

Product: (αβ)³ = (-4)³ = -64

Equation: x² - (sum)x + product = 0

So: x² - 24x - 64 = 0

🔢 Problem 4: Find 'a' in 7x² + ax + 2 = 0

Given roots α and β with β - α = -13/7

We know:

α + β = -a/7

αβ = 2/7

And β - α = -13/7

Square both sides of (β - α):

(β - α)² = (-13/7)² ⇒ β² - 2αβ + α² = 169/49

But α² + β² = (α+β)² - 2αβ, so:

(α+β)² - 4αβ = 169/49

Substitute known values:

(-a/7)² - 4(2/7) = 169/49

a²/49 - 8/7 = 169/49

Multiply all by 49:

a² - 56 = 169 ⇒ a² = 225 ⇒ a = ±15

But we need to check which solution fits:

If a = 15: α+β = -15/7, β-α = -13/7 ⇒ β = -2, α = 1/7 (valid)

If a = -15: α+β = 15/7, β-α = -13/7 ⇒ β = 1/7, α = 2 (valid)

Both solutions are possible: a = 15 or a = -15

🎯 Problem 5: Find 'a' in 2y² - ay + 64 = 0

Given one root is twice the other. Let roots be k and 2k.

Sum of roots: k + 2k = 3k = a/2 ⇒ a = 6k

Product of roots: k × 2k = 2k² = 64/2 = 32 ⇒ k² = 16 ⇒ k = ±4

Case 1: k = 4 ⇒ a = 6×4 = 24

Case 2: k = -4 ⇒ a = 6×(-4) = -24

Possible values: a = 24 or a = -24

🌟 Problem 6: Find 'k' in 3x² + kx + 81 = 0

Given one root is square of the other. Let roots be m and m².

Product of roots: m × m² = m³ = 81/3 = 27 ⇒ m = 3

So roots are 3 and 9.

Sum of roots: 3 + 9 = 12 = -k/3 ⇒ k = -36

Check if m = -3:

Then roots would be -3 and 9, but product would be -27 ≠ 27.

Only valid solution: k = -36