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October 21st, 2017, 11:34 PM   #1
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Factoring

A yacht sails a leg of a race by going in a straight course from point A to point B for (x^2 + 8x) km. From point B it travels to point C for (2x+11)km, then back to point A for (4x+24)km.
a) write a simplified expression in km for the total distance the yacht sailed
b) factorise the expression in (a)
c) the factorised expression also represents the product of the length and width of a rectangle, and if the length is 15km, find the value of x. Give your answer as an exact distance
d) Find the length of the race, rounded to two decimal places, by substituting the exact value for x into the expression for the total distance in (a)

My answer:
a) (x^2 + 8x) + (2x+11) + (4x+24) = x^2 + 14x + 35
b) I can't work out how to factorise (x^2 + 14x + 35). Any help would be appreciated.
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October 21st, 2017, 11:53 PM   #2
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b) you can always use the quadratic formula

$\begin{align*}

r_1, r_2 &= \dfrac{-14 \pm \sqrt{14^2 - (4)(35)}}{2} \\

&= \dfrac{-14 \pm \sqrt{56}}{2} \\

&= -7 \pm \sqrt{14}

\end{align*}$

so

$x^2 + 14x + 35 = (x-r_1)(x-r_2) = (x + 7 - \sqrt{14})(x + 7 + \sqrt{14})$
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October 22nd, 2017, 12:10 AM   #3
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Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
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October 22nd, 2017, 12:20 AM   #4
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Quote:
Originally Posted by pianist View Post
Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
I suppose if you are really good you can "foil" the thing by eye even though the roots aren't rational.

The quadratic formula is the way to go when the factorization isn't immediately apparent.
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October 23rd, 2017, 11:33 AM   #5
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Quote:
Originally Posted by pianist View Post
Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
You get the "quadratic formula" itself by applying "completing the square" to the general quadratic $\displaystyle ax^2+ bx+ c= 0$.

Here, the quadratic expression is $\displaystyle x^2+ 14x+ 35$. Knowing that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$ and comparing those two, 2a= 14 so a= 7, $\displaystyle a^2= 49$. $\displaystyle x^2+ 14x+ 35= x^2+ 14x+ 49- 49+ 35= (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2= (x+ 7+ \sqrt{14})(x+ 7- \sqrt{14})$.
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Last edited by skipjack; October 24th, 2017 at 06:12 AM.
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October 24th, 2017, 05:02 AM   #6
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Quote:
Originally Posted by Country Boy View Post
You get the "quadratic formula" itself by applying "completing the square" to the general quadratic $\displaystyle ax^2+ bx+ c= 0$.

Here, the quadratic expression is $\displaystyle x^2+ 14x+ 35$. Knowing that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$ and comparing those two, 2a= 14 so a= 7, $\displaystyle a^2= 49$. $\displaystyle x^2+ 14x+ 35= x^2+ 14x+ 49- 49+ 35= (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2= (x+ 7+ \sqrt{14})(x+ 7- \sqrt{14})$.
Two typos in this. First, "$\displaystyle a^2= 49$" should have been "$\displaystyle a^2= 4$".

More importantly, "$\displaystyle (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2$" should have been "$\displaystyle (x+ 7)^2- 14= (x+ 7)^2- (\sqrt{14})^2$". Then we can use "$\displaystyle a^2- b^2= (a+ b)(a- b)$.

Last edited by skipjack; October 24th, 2017 at 06:13 AM.
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