Question

3. Show that the roots of the equation

Answer 1

Steps :
1) Here,
${x}^{2} + 2(a + b)x +2 ( {a}^{2} + {b}^{2} ) = 0$
Here,
Discriminant,
D =
${(2(a + b))}^{2} - 4 \times 2 \times ( {a}^{2} + {b}^{2} ) \\ = > 4 ({a}^{2} + 2ab + {b}^{2} ) - 8( {a}^{2} + {b}^{2} ) \\ = > - 4( {a}^{2} - 2ab + {b}^{2} ) \\ = > - 4 {(a - b)}^{2}$

2) Here,
D < 0 if a is not equal to b.
So, if a is not equal to b, then
Roots of the given quadratic equation are imaginary or not real.

Answer 2

x² + 2(a + b)x + 2(a² + b²) = 0.
where, a = 1 , b = 2(a + b) , c = 2(a² + b²)

D = b² - 4ac
D = [2(a + b)]² - 4(1)[2(a² + b²)]
D = 4a² + 4b² + 8ab - 8a² - 8b²
D = -4a² - 4b² + 8ab
D = -4(a + b)²
so , we know if D < 0 hence,the roots are unequal and imaginary roots.