Question

show that the root of the equation x²+2(a+b)x+2(a²+b²)=0 are not real

Answer 1

Given :

x²+2(a+b)+2(a²+b²) =0 is a quadratic equation.

To prove :

The roots of x²+2(a+b)x+2(a²+b²)=0 are not real.

Proof :

A quadratic equation has the highest degree as 2. And is expressed in the form ax²-bx +c.

Some of the discriminant of roots are :

b²- 4ac = 0 [Rational & equal roots]

b²- 4ac > 0 [ Real but unequal roots]

b²- 4ac < 0 [ non - real roots]

So, for non-real roots

Here,

a = 1

b = 2(a+b)

c = 2(a²+b²)

Now,

b ² - 4ac < 0

[2(a+b)]² - 4[1 × 2(a²+b²)] < 0

4(a²+2ab+b²) - 8(a²+b²) < 0

4a² + 8ab +4b² - 8a² - 8b² < 0

-4a²- 4b² +8ab < 0

-a ² - b ² +2ab < 0

a ² + b ² - 2ab < 0

(a-b) ²< 0

Therefore, from this we can conclude that if (a-b) ² is less than 0, then the roots of quadratic equation x²+2(a+b)x+2(a²+b²)=0 are not real.

Answer 2

$\huge\red{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{x^{2}+2(a+b)x+2(a^{2}+b^{2})=0}}$

$\sf\pink{To \ show:}$

$\sf{Roots \ are \ not \ real.}$

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

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{x^{2}+2(a+b)x+2(a^{2}+b^{2})=0}}$

$\sf{Here,}$

$\sf{a=1}$

$\sf{b=2(a+b)}$

$\sf{c=2(a^{2}+b^{2})}$

$\sf{Discriminant=b^{2}-4ac}$

$\sf{\Delta=[2(a+b)]^{2}-4(1)[2(a^{2}+b^{2})]}$

$\sf{\Delta=4a^{2}+4ab+b^{2}-4(2a^{2}+2b^{2})}$

$\sf{\Delta=4a^{2}+4ab+b^{2}-8a^{2}-8b^{2}}$

$\sf{\Delta=-4a^{2}+4ab-4b^{2}}$

$\sf{\Delta=-4(a^{2}-ab+b^{2})}$

$\sf{\Delta=-4(a-b)^{2}}$

$\sf{But, \ it's \ in \ negative}$

$\sf{\therefore{\Delta<0}}$

$\sf\purple{\tt{\therefore{Roots \ are \ not \ real.}}}$