$$ x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$$
I tried to make an inequality using the discriminant and I simplified it to get $$a^2+2ab+b^2+\frac{1}{2}≤0$$
But I don't know how to solve this.
2026-04-03 04:53:17.1775191997
If this equation in x has real roots find the value of a and b.
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This equation has a real root then the discriminant must be non-negative.
$$\Delta=4(1+a)^2-4(3a^2+4ab+4b^2+2) \geq 0$$ $$\iff 3a^2+4ab+4b^2-a^2-2a+1 \leq 0$$ $$ \iff (a-1)^2+(a+2b)^2 \leq 0$$ $$\iff a-1=a+2b=0$$ $$\iff a=1, b=-\frac{1}{2}$$