Maths Olympiad Prep

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Algebra Difficulty 1.4 Junior Find the answer Canada

If the equation $(x+2)(x+t)= x^2 + bx
+ 12 is true for all real numbers xx, the value of b$ is

Pick one

Solution

Since the equation is true for all real numbers xx, it is true for x=0x=0.

Substituting x=0x=0, we get (0+2)(0+t)=02+b(0)+12(0+2)(0+t)=0^2+b(0)+12 or 2t=122t=12, and so t=6t=6.

The equation becomes (x+2)(x+6)=x2+bx+12(x+2)(x+6)=x^2+bx+12 or x2+8x+12=x2+bx+12x^2+8x+12=x^2+bx+12, and so b=8b=8.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.