Question

show that thw eqn 2(a^2+b^2)x^2+2(a+b)x+1=0 has no real roots

Answer 1

Answer:

Step-by-step explanation:

On comparing with Ax2 + Bx + C = 0

We get A = 2(a2 +b2), B = 2(a + b), C = 1

Discriminant of the quadratic equation = B2 – 4AC

Answer 2

Answer:

Step-by-step explanation:

$2(a^2+b^2)x^2+2(a+b)x+1=0$

For Quadratic Equation $ax^{2}+bx+c=0 \text{ [where x is the variable and a, b and c are known values]}$

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

$b^{2}-4ac \text{ is called Discriminant (D)}$

when Discriminant (D) is positive, we get two Real solutions  for x

when Discriminant (D) is zero we get just ONE real solution (both answers are the same)

when Discriminant (D) is negative we get a pair of Complex solutions

In this equation the Discriminant (D) is

$D=b^{2}-4ac\\\\\Rightarrow D=[2(a+b)]^2-4\times[2(a^2+b^2)]\times1\\\\\Rightarrow D= (2a+2b)^2-8(a^2+b^2)\\\\\Rightarrow D= (4a^2+8ab+4b^2)-8a^2-8b^2\\\\\Rightarrow D= (-4a^2+8ab-4b^2)\\\\\Rightarrow D= -4(a^2-2ab+b^2)\\\\\Rightarrow D= -4(a-b)^2$

The value of $(a-b)^2>0$

Therefore the value of D is always negative.

when Discriminant (D) is negative we get a pair of Complex solutions