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:

Given Equation: 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

Answer 2

Answer:-

X² -6x + 11 = 0

Explanation:-

Given:-

α and β are the zeros of the equation,

x²-2x+3 = 0

To Find:-

The equation, whose roots are α+2 and β+2

Solution:-

Let, us find α and β

x²-2x+3x = 0

Sum of the roots

sum of the roots = $\dfrac{-b}{a}$

$\alpha +\beta = \dfrac{-(-2)}{1}$

$\alpha +\beta = 2$

Product of the roots

product of the roots = $\dfrac{c}{a}$

$\alpha \beta = \dfrac{3}{1}$

$\alpha \beta = 3$

Now, roots of new equatio. are α+2, β+2.

Sum of the roots of new equation are

$\implies \alpha+2+ \beta +2$

$\implies \alpha+ \beta +4$

$\implies 2+4$

$\implies 6$

Product of the roots of new equation:

$\implies( \alpha+2) (\beta +2)$

$\implies \alpha \beta +2\beta + 2\alpha +4$

$\implies \alpha \beta +2\beta + 2\alpha +4$

$\implies 3 + 2(\alpha+\beta )+4$

$\implies 3 + 2(2)+4$

$\implies 3 + 4 + 4$

$\implies 11$

Now,

Using quadratic formula,

$= {x} ^{2}-[(\alpha+2)+(\beta+2)]x+(\alpha+2)( \beta+2)$

$= {x} ^{2}-[6]x+11$

$= {x} ^{2}-6x+11$

Hence, the required equation is

${x} ^{2}-6x+11$