已知实数a,b满足(a²+b²)²-2(a²+b²)=8,则a²+b²的值为
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![已知实数a,b满足(a²+b²)²-2(a²+b²)=8,则a²+b²的值为](/uploads/image/z/603087-15-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0a%2Cb%E6%BB%A1%E8%B6%B3%EF%BC%88a%26%23178%3B%2Bb%26%23178%3B%EF%BC%89%26%23178%3B-2%EF%BC%88a%26%23178%3B%2Bb%26%23178%3B%EF%BC%89%3D8%2C%E5%88%99a%26%23178%3B%2Bb%26%23178%3B%E7%9A%84%E5%80%BC%E4%B8%BA)
已知实数a,b满足(a²+b²)²-2(a²+b²)=8,则a²+b²的值为
已知实数a,b满足(a²+b²)²-2(a²+b²)=8,则a²+b²的值为
已知实数a,b满足(a²+b²)²-2(a²+b²)=8,则a²+b²的值为
(a²+b²)²-2(a²+b²)=8
令(a²+b²)=z>0
则 z²-2z-8=0
(z-4)(z+2)=0
z=4,z=-2(舍去)
所以(a²+b²)=z=4
(a²+b²)²-2(a²+b²)=8
(a²+b²)²-2(a²+b²)-8=0
【(a²+b²)-4】【(a²+b²)+2】=0
所以(a²+b²)-4=0或(a²+b²)+2=0、
a²+b²=4
取t=a²+b²,则t>0.
(a²+b²)²-2(a²+b²)=8化为t²-2t-8=0,即(t+2)(t-4)=0
解得:t=-2或t=4
又t>0,所以则t=4,即a²+b²=4