Question

If the ratio of $x^2+bx+c =0$ and $x^2+qx+r =o$ are same , then
Options :-
$r^2c= qb^2 \\r^2b = qc^2 \\rb^2 = cq^2 \\rc^2 = bq^2$

Answer 1

Given :-

If  the ratio of  x² + bx + c = 0 and x² + qx + r = 0 are same , then

To Find :-

Relationship between them

Solution :-

Let the roots of

x² + bx + c = 0 be α and β

and,

x² + qx + r = 0 be α' and β'

Accoding to the statement

α + β/α - β = α' + β'/α' + β'

Sum of zeroes = -b/a

Sum = -(b)/1

Sum = -b

Product = c/a

Product = c/1

Product = c

Now, In Eq 2

Sum = -b/a

Sum = -(q)/1

Sum = -q

Product = c/a

Product = r/1

Product = r

α - β = √(b² - 4ac)

-b/√(b² - 4(1)(c) = -q/√(q² - 4(1)(r)

-b/√(b² - 4c) = -q/√(q² - 4r)

Now, On squaring both side we get

[-b/√(b² - 4c)]² = [-q/√(q² - 4r)]²

b²/b² - 4c = q²/q² - 4r

Cross multiplication

b²(q² - 4r) = q²(b² - 3c)

-4b²r = -4cq²

Cancel -4 from both side

b²r = cq²

b²r can be written as rb²

rb² = cq²

Therefore, Option C is correct

Answer 2

Answer :-

Correct relationship is rb² = cq² . [Option.3] .

Explanation :-

If ratio of roots of the two quadratic equations are same, then the solution is .

We have :-

→ 1st equation : x² + bx + c = 0

→ 2nd equation : x² + qx + r = 0

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For 1st equation :-

Let the roots be α₁ and β₁ .

⇒ Sum of roots = -b/1

⇒ α₁ + β₁ = -b/1

⇒ α₁ + β₁ = -b

⇒ Product of roots = c/1

⇒ α₁β₁ = c

By relationship (a - b)² = (a + b)² - 4ab :-

⇒ (α₁ - β₁)² = (α₁ + β₁)² - 4α₁β₁

⇒ (α₁ - β₁)² = (-b)² - 4c

⇒ α₁ - β₁ = √(b² - 4c) ---(1)

For 2nd equation :-

Let the roots be α₂ and β₂ .

⇒ Sum of roots = -q/1

⇒ α₂ + β₂ = -q

⇒ Product of roots = r/1

⇒ α₂β₂ = r

Again by same identity used above :-

⇒ (α₂ - β₂)² = (α₂ + β₂)² - 4α₂β₂

⇒ (α₂ - β₂)² = (-q)² - 4r

⇒ α₂ - β₂ = √(q² - 4r) ----(2)

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Now, as we have α₁ : β₁ = α₂ : β₂ , so by using Componendo and Dividendo rule :-

⇒ (α₁ + β₁)/(α₁ - β₁) = (α₂ + β₂)/(α₂ - β₂)

⇒ -b/√(b² - 4c) = -q/√(q²- 4r)

⇒ [-b/√(b² - 4c)]² = [-q/√(q² - 4r)]²

⇒ b²/(b² - 4c) = q²/(q² - 4r)

⇒ b²(q² - 4r) = q²(b² - 4c)

⇒ -4b²r = -4cq²

⇒ rb² = cq²