Question
There are three consecutive integers such that the square of the first increased by the product of the other two gives $154$. What are the integers.
✓
Answer
Let first integer $= x$
Then second integer $=x+1$, and third integer $=x+2$
According to the condition,
$\Rightarrow x^2+(x+1)(x+2)=154$
$\Rightarrow x^2+x^2+3 x+2=154$
$\Rightarrow 2 x^2+3 x+2-154=0$
$\Rightarrow 2 x^2-16 x+19 x-152=0$
$\Rightarrow 2 x(x-8)+19(x-8)=0$
$\Rightarrow(x-8)(2 x+9)=0$
Either $x-8=0$, then $x=8$
or $2 x+19=0$, then $2 x=-19$
$\Rightarrow x=\frac{-19}{2}$ But it is not an integer.
First number $=8$
Second number $=8+1=9$ and third number $=8+2=10$
Then second integer $=x+1$, and third integer $=x+2$
According to the condition,
$\Rightarrow x^2+(x+1)(x+2)=154$
$\Rightarrow x^2+x^2+3 x+2=154$
$\Rightarrow 2 x^2+3 x+2-154=0$
$\Rightarrow 2 x^2-16 x+19 x-152=0$
$\Rightarrow 2 x(x-8)+19(x-8)=0$
$\Rightarrow(x-8)(2 x+9)=0$
Either $x-8=0$, then $x=8$
or $2 x+19=0$, then $2 x=-19$
$\Rightarrow x=\frac{-19}{2}$ But it is not an integer.
First number $=8$
Second number $=8+1=9$ and third number $=8+2=10$
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