Question

Find the quadratic equation whose roots exceed the roots of the quadratic equation x2-2x+3 by 2​

Answer 1

Answer:

x² - 6x + 11

Step-by-step explanation:

Let f(x) = x² - 2x + 3.

Suppose a and b are the roots, so f(a) = f(b) = 0.

We want a polynomial g(x) that has roots a+2 and b+2.

So we want

g(a+2) = g(b+2) = 0.          ... (*)

Take g(x) = f(x-2).

Then g(a+2) = f( (a+2) - 2 ) = f(a) = 0

and g(b+2) = f( (b+2) - 2 ) = f(b) = 0.

So this is the quadratic that satisfies (*).

The quadratic that we want is then

g(x) = f(x-2) = (x-2)² - 2(x-2) + 3

= x² - 4x + 4 - 2x + 4 + 3

= x² - 6x + 11

Answer 2

Answer:

VERY EASY METHOD based on the syllabus.

Step-by-step explanation:

x²-2x+3=0

Let m and n be the roots of the given equation.

Therefore, sum of roots=

m + n = -b/a

= -(-2)/1

= 2 .........( 1 )

Let m+2 and n+2 be the roots of the new equation.

Hence, sum of new roots =

m + 2 + n + 2 = m + n + 4

=2 + 4 ........( from 1 )

=6

Products of roots of the equation =

mn = c/a

=3/1

=3

Product of roots = ( m + 2 ) ( n + 2 )

=m (n+2) + 2 (n + 2)

= mn + 2m + 2n + 4

= mn +2(m + n) + 4

=3 + 2X2 + 4

=3+8

= 11

Formula

x² - (m+n)x + MN = 0

=x² - 6x+ 11=0