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I have been a bit stuck on this question.

The product of two consecutive odd numbers is $143$. Find the next numbers.

I have made this into: $$ (2x+1)(2x+3)=143. $$ I got $x_1 = -7$ and $x_2 = -5$ for this, which doesn't seem right.

How can I fix this? (By the way we are only able to solve for $x$ by using factorising and not with the quadratic formula).

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    $\begingroup$ You should get -7 and 5. Alternatively you can just choose the easier set up with: x(x+2)= 143 and choosing the value of x which is odd. $\endgroup$ – thedilated May 24 '16 at 6:58
  • $\begingroup$ -7 and 5 sorry. But what would I do after this? $\endgroup$ – Christopher U'Ren May 24 '16 at 6:59
  • $\begingroup$ Once you get 5. You can substitute back into 2(5) +1 and 2(5)+3 which gives you 11 and 13 as the two odd numbers. $\endgroup$ – thedilated May 24 '16 at 7:00
  • $\begingroup$ It's easier if you express the equation as $(2x-1)(2x+1)=143$. $\endgroup$ – John Joy May 24 '16 at 13:41
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\begin{align*} (2x + 1)(2x + 3) & = 143\\ 2x(2x + 3) + 1(2x + 3) & = 143 && \text{expand}\\ 4x^2 + 6x + 2x + 3 & = 143 && \text{apply the distributive law}\\ 4x^2 + 8x + 3 & = 143 && \text{combine like terms}\\ 4x^2 + 8x - 140 & = 0 && \text{subtract $140$ from each side of the equation}\\ x^2 + 2x - 35 & = 0 && \text{divide each side of the equation by $4$}\\ (x + 7)(x - 5) & = 0 && \text{factor the quadratic} \end{align*} If a product is equal to zero, then one of the factors must be equal to zero. Hence, \begin{align*} x + 7 & = 0 & x - 5 & = 0\\ x & = -7 & x & = 5 \end{align*} You can verify that these values are correct by substituting them into the equation $(2x + 1)(2x + 3) = 143$.

If $x = -7$, then the consecutive odd numbers are $2(-7) + 1 = -13$ and $2(-7) + 3 = -11$.

If $x = 5$, then the consecutive odd numbers are $2(5) + 1 = 11$ and $2(5) + 3 = 13$.

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An idea: decompose in primes product $\;143=11\cdot13\;$ .

Observe that we're given $\;(2x+1)(2x+3)=143\;$ , and the left side is the product of two consecutive odd integers. Once you get the prime product for $\;143\;$ you already have not many choices: one must be $\;11\;$ , and the other one $\;13\;$ ...or both negative: $\;-11,\,-13\;$ , and you get:

$$\begin{cases}2x+1=11\\2x+3=13\end{cases}\implies \color{red}{x=5}\;,\;\;\begin{cases}2x+1=-13\\2x+3=-11\end{cases}\implies \color{red}{x=-7}$$

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$$(2x + 1)(2x + 3) = 143$$ Let $y=\frac{(2x+1)+(2x+3)}{2}=2x+2$. Then: $$(y-1)(y+1)=143\implies y^2-1=143=144-1$$ Thus, $y^2=144\implies y=\pm12\implies(2x+2)=\pm12\implies x\in\{-7,5\}$

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