If a and b are distinct real numbers, show that the quadratic equation
![2(a {}^{2} + b {}^{2} )x {}^{2} + 2(a + b)x + 1 = 0 \\ \\ has \: \:no \: \: real \: \: roots. 2(a {}^{2} + b {}^{2} )x {}^{2} + 2(a + b)x + 1 = 0 \\ \\ has \: \:no \: \: real \: \: roots.](https://tex.z-dn.net/?f=2%28a+%7B%7D%5E%7B2%7D++%2B+b+%7B%7D%5E%7B2%7D+%29x+%7B%7D%5E%7B2%7D++%2B+2%28a+%2B+b%29x+%2B+1+%3D+0+%5C%5C++%5C%5C+has+%5C%3A++%5C%3Ano+%5C%3A++%5C%3A++real+%5C%3A++%5C%3A+roots.)
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discriminant
= 4( a+b)^2 -8(a^2 + b^2)
4 a^2+ 4 b^2 +8 ab - 8 a^2 - 8 b^2
- 4 a^2 - 4b^2 + 8 ab
- ( 2a -2b)^2 <0
So it has no real roots
= 4( a+b)^2 -8(a^2 + b^2)
4 a^2+ 4 b^2 +8 ab - 8 a^2 - 8 b^2
- 4 a^2 - 4b^2 + 8 ab
- ( 2a -2b)^2 <0
So it has no real roots
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