Question

A student noticed that the roots of the equation x² + bx + a = 0 are each 1 less than the roots of the equation x² + ax + b = 0. Then a + b = ?

Answer 1

If  δ and Ψ are the roots of Quadratic Equation px² + qx + r = 0 then :

✿  Sum of the roots : δ + Ψ $\bf{= \frac{-q}{p}}$

✿  Product of the roots : δ × Ψ $\bf{= \frac{r}{p}}$

Let the Roots of the Equation x² + ax + b = 0 be 'α' and 'β'

⇒ Sum of the roots : α + β = -a

⇒ Product of the roots : α × β = b

Given : The Roots of the Equation x² + bx + a = 0 are each 1 less than the roots of the Equation x² + ax + b = 0

⇒ The Roots of the Equation x² + bx + a = 0 are (α - 1) and (β - 1)

⇒ Sum of the Roots : (α - 1) + (β - 1) = -b

⇒ (α + β) - 2 = -b

But, We know that : (α + β) = -a

⇒ (-a - 2) = -b

⇒ a + 2 = b

⇒ a - b = -2 --------- [1]

⇒ Product of the Roots : (α - 1)(β - 1) = a

⇒ αβ - (α + β) + 1 = a

But, We know that : α × β = b and α + β = -a

⇒ b + a + 1 = a

⇒ b = -1

Substituting b = -1 in Equation [1], We get :

⇒ a + 1 = -2

⇒ a = -3

⇒ a + b = (-3 - 1) = -4

Answer 2

Answer:

a + b = -4

Step-by-step explanation:

Let the roots of the equation is α,β.

(1)

Given quadratic equation is x² + bx + a = 0.

Here, a = 1, b = b, c = a

(i)

Sum of roots = -b/a

α + β = -b.

(ii)

Product of roots = c/a

αβ = a

(2)

Given Equation is x² + ax + b = 0.

Here, a = 1, b = a, c = b

Given that roots are each 1 less than the roots of equation x² + ax + b.

(iii)

Sum of roots = -b/a

(α - 1) + (β - 1) = -a

(iv)

Product of roots = c/a

(α - 1)(β - 1) = b

αβ - (α + β) + 1 = b

a - (-b) + 1 = b

a + b + 1 = b

a + 1 = 0

a = -1.          --------- (iv)

Substitute a = -1 in (iii), we get

⇒ α - 1 + β - 1 = -a

⇒ - b- 2 = -a

⇒ -(b + 2) = -a

⇒ a = b + 2         ------ (v)

Substitute (iv) in (v), we get

⇒ -1 = b + 2

⇒ b = -3.

Hence:

⇒ a + b

⇒ -1 - 3

⇒ -4.

Therefore,a + b = -4.

Hope it helps!