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प्रश्न
Show that the equation 2(a2 + b2)x2 + 2(a + b)x + 1 = 0 has no real roots, when a ≠ b.
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उत्तर
The quadric equation is 2(a2 + b2)x2 + 2(a + b)x + 1 = 0
Here,
a = 2(a2 + b2), b = 2(a + b) and c = 1
As we know that D = b2 - 4ac
Putting the value of a = 2(a2 + b2), b = 2(a + b) and c = 1
D = {2(a + b)}2 - 4 x (2(a2 + b2)) x (1)
= 4(a2 + 2ab + b2) - 8(a2 + b2)
= 4a2 + 8ab + 4b2 - 8a2 - 8b2
= 8ab - 4a2 - 4b2
= -4(a2 - 2ab + b2)
= -4(a - b)2
We have,
a ≠ b
a - b ≠ 0
Thus, the value of D < 0
Therefore, the roots of the given equation are not real
Hence, proved
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