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 January 5th, 2017, 01:51 AM #1 Newbie   Joined: Dec 2016 From: nottinghamshire,England Posts: 2 Thanks: 0 Quadratic by factorising Hi All I need help with the method (not the answer) how do I got about solving 10x^2+3x-4=0 I need to solve the following quadratic by factorising thank you
 January 5th, 2017, 02:20 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,519 Thanks: 506 Math Focus: Yet to find out. Many ways. You could start by dividing through by 10. $10x^2 + 3x - 4 = 0$ Divide through by 10, $x^2 + \frac{3}{10} x - \frac{4}{10} = 0.$ Then ask yourself, what two numbers add together to give $\frac{3}{10}$, and multiply together to give $\frac{4}{10}$. With a bit of fiddling you should find that, $- \frac{1}{2} + \frac{4}{5} = \frac{3}{10}$ and that $- \frac{1}{2} * \frac{4}{5} = - \frac{4}{10}.$ Finally we have, $(x - \frac{1}{2})(x + \frac{4}{5}) = 0.$ And so, the solutions for $\displaystyle x$ are, $x = \frac{1}{2},$ and $x = - \frac{4}{5}.$
 January 5th, 2017, 06:26 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,701 Thanks: 1359 Some good advice here ... Factoring Trinomials When the Leading Coefficient is not 1
January 5th, 2017, 05:05 PM   #4
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 Originally Posted by claire123 Hi All I need help with the method (not the answer) how do I got about solving 10x^2+3x-4=0 I need to solve the following quadratic by factorising thank you
Let's start by expanding out \displaystyle \begin{align*} \left( x + m \right) \left( x + n \right) \end{align*}, where m and n are just constants. Then we get

\displaystyle \begin{align*} \left( x + m \right) \left( x + n \right) &= x \left( x + n \right) + m \left( x + n \right) \\ &= x^2 + n\,x + m\,x + m\,n \\ &= x^2 + \left( m + n \right) x + m\,n \end{align*}

which is now a monic quadratic of the form \displaystyle \begin{align*} x^2 + b\,x + c \end{align*}, where \displaystyle \begin{align*} b = m + n \end{align*} and \displaystyle \begin{align*} c = m\,n \end{align*}. So if you wanted to go backwards, you need to find the right two numbers (m and n) which multiply to give b and add to give c.

Things get a little trickier when the quadratic is not monic though. Let's look at what happens when we expand \displaystyle \begin{align*} \left( m\,x + n \right) \left( p\,x + q \right) \end{align*} (here m, n, p, q are just constants). It also helps to remember that any combination of them is also just a constant. Anyway, we get

\displaystyle \begin{align*} \left( m\,x + n \right) \left( p\,x + q \right) &= m\,x \left( p\,x + q \right) + n \left( p\,x + q \right) \\ &= m\,p\,x^2 + m\,q\,x + n\,p\,x + n\,q \\ &= m\,p\,x^2 + \left( m\,q + n\,p \right) x + n\,q \end{align*}

which is now of the form \displaystyle \begin{align*} a\,x^2 + b\,x + c \end{align*} where \displaystyle \begin{align*} a = m\,p , \, b = m\,q + n\,p , \, c = n\,q \end{align*}.

Here it is not as simple as the monic case, because the middle term depends on four constants. But what we can do is multiply the a and c terms together. This gives

\displaystyle \begin{align*} a \cdot c &= m\,p \cdot n\,q \\ &= m\,q \cdot n\,p \end{align*}

Now as I said before, any combination of the four constants is also just a constant. So here, when we have multiplied the a and c values together, it again gives the same combination of two numbers (mp and nq) multiplied together as the middle term does when they're added. So we can still use the "times to give, add to give" process, as long as we have first multiplied the a and c terms.

Let's look at your example \displaystyle \begin{align*} 10\,x^2 + 3\,x - 4 \end{align*}. We need to multiply a and c, which is \displaystyle \begin{align*} 10 \left( -4 \right) = -40 \end{align*}.

So what we want are two numbers that multiply to give -40 (as they multiply to give a negative number, we know they need different signs) and add to give 3 (well really have a difference of 3 as we know that they have to have different signs - also the larger of the two numbers will be the positive).

The two numbers are 8 and -5. So now we break the middle term up as the sum of these two numbers and continue to factorise...

\displaystyle \begin{align*} 10\,x^2 + 3\,x - 4 &= 10\,x^2 + \left( 8 - 5 \right) x - 4 \\ &= 10\,x^2 + 8\,x - 5\,x - 4 \\ &= 2\,x\left( 5\,x + 4 \right) - 1 \left( 5\,x + 4 \right) \\ &= \left( 5\,x + 4 \right) \left( 2\,x - 1 \right) \end{align*}

 January 6th, 2017, 03:25 AM #5 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 Use quadratic formula to factor out it, $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ Put values in formula $\Rightarrow \dfrac{-3\pm \sqrt{3^2-4×10(-4)}}{2×10}$ $\Rightarrow \dfrac{-3\pm \sqrt{169}}{20}$ $\Rightarrow \dfrac{-3\pm 13}{20}$ $x=-\dfrac{4}{5}$ $x=\dfrac{1}{2}$

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