Question

find the quadratic equation whose roots exceed the roots of the quadratic equation x2-2x+3=0 by 2​

Answer 1

Answer:

x=2

Step-by-step explanation:

=2(2)-2(2)+3

=4-4+3

=0+3

=3

Answer 2

The  quadratic equation

$x^2-6x+11$

Step-by-step explanation:

Given quadratic equation

$x^2-2x+3=0$

Compare it with

$ax^2+bx+c=0$

$a=1,b=-2,c=3$

Quadratic formula :

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Using the formula

$x=\frac{2\pm\sqrt{(-2)^2-4(1)(3)}}{2}$

$x=\frac{2\pm\sqrt{4-12}}{2}$

$x=\frac{2\pm i2\sqrt 2}{2}$

$x=\alpha=\frac{2+i2\sqrt 2}{2}=2\times \frac{1+i\sqrt 2}{2}=1+i\sqrt 2$

$x=\beta=\frac{2-i2\sqrt 2}{2}=2\times \frac{1-i\sqrt 2}{2}=1-i\sqrt 2$

Let p and q are roots of quadratic equation $x^2-(p+q)x+pq$

$p=1+i\sqrt 2+2=3+i\sqrt 2$

$q=1-i\sqrt 2+2=3-i\sqrt 2$

Substitute the values

$x^2-(3+i\sqrt 2+3-i\sqrt 2)x+(3+i\sqrt 2)(3-i\sqrt 2)$

$x^2-6x+(3^2-(i\sqrt 2)^2)$

Using identity:$a^2-b^2=(a+b)(a-b)$

$x^2-6x+(9+2)=x^2-6x+11$

By using $i^2=-1$

Hence, the  quadratic equation

$x^2-6x+11$

#Learns more:

https://brainly.in/question/8635897:Answered by Bhartisurana