The number of real roots of the equation (x+1)^2+|x-5|=274 is Answer: 2

Question

The number of real roots of the equation $(x+1)^2+|x-5|=\dfrac{27}4$ is

Answer: 2

Accepted Answer

Case $x\ge5$: $(x+1)^2+(x-5)=	frac{27}4 \implies 4x^2+12x-43=0$, both roots come out less than 5, so no solution here.

Case $x<5$: $(x+1)^2-(x-5)=	frac{27}4 \implies 4x^2+4x-3=0 \implies x=\dfrac{-4\pm8}{8}=\dfrac12,\ -\dfrac32$, both less than 5 --- valid.

Hence there are 2 real roots.

Solutions · 1