Question

if alpha and beta are root of the equation x^2-2x+3=0 then the equation whose roots are alpha+2 and beta +2​

Answer 1

Answer:

Step-by-step explanation:

Let α, β  be the roots of the equation

x² - 2x + 3 = 0,

Let f(x) = x² - 2x + 3

Given that roots are increased by 2, so new roots

of the equation are α + 2, β + 2

Let y = α + 2 which is the required root of new

equation,

So, α = y - 2

But, we know α is root of f(x), hence

f(α) = 0

But α = y - 2, so

f(y - 2) = 0

(y - 2)² - 2(y - 2) + 3 = 0

y² - 6y + 11 = 0,

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

Given Equation:

$\boxed{x² - 2x + 3 = 0}$

Sum of roots = α + β = -b/a = -(-2)/1 = 2

⇒ Sum of roots of new equation = α + 2 + β + 2

⇒ Sum of roots of new equation = α + β + 4

⇒ Sum of roots of new equation = 2 + 4 = 6

Similarly, Product of roots = c/a = αβ = 3/1 = 3

Product of roots of new equation = ( α + 2 ) ( β + 2 )

⇒ Product of roots of new equation = αβ + 2α + 2β + 4

⇒ Product of roots of new equation = αβ + 2 ( α + β ) + 4

⇒ Product of roots of new equation = 3 + 2 ( 2 ) + 4

⇒ Product of roots of new equation = 3 + 4 + 4 = 11

Therefore new equation is given as:

⇒ x² - ( Sum of roots ) x  + Product of roots

⇒ x² - 6x + 11 = 0 Required Equation