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How to solve quadratic equations - Factor quadratic equations

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How to solve quadratic equations - Factor quadratic equations  [#permalink]

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New post 27 May 2017, 09:42
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How to solve quadratic equations - Factor quadratic equations

Edit: For solving quadratic inequation, you may find this useful tool: Factor table with sign


Any quadratic equations have the same form like this \(ax^2+bx+c=0\)

Now, let's start with these questions.

Question 1. Solve \(x^2 + 6x +5 = 0\)

Question 2. Solve \(x^2 + 4x -12 =0\)


How did you solve those quetsions? :-D You may see the simple solution like these:
\(x^2 + 6x +5 = 0 \iff (x+1)(x+5)=0 \iff x=-1\) or \(x=-5\)
\(x^2 + 4x -12 = 0 \iff (x+6)(x-2)=0 \iff x=-6\) or \(x=2\)

Have you ever asked how to come to these solution? Now, I'll guide you how :-D

Approach 1. Factor quadratic equations by factorization

Let's come back with Question 1.
Note that \(6=2\times 3\) and \(2+3=5\). Hence we have \(x^2+5x+6=(x+2)(x+3)\).

This approach comes from the theory that \((x+m)(x+n)=x^2+(m+n)x+mn \implies b=m+n\) and \(c=mn\).

How about Question 2?
Easily, we have \(-12=(-3) \times 4 = 3 \times (-4)=(-2) \times 6 = 2 \times (-6) = 1 \times 12 = (-1) \times 12\).

We choose \(12=(-2) \times 6\) because \((-2) + 6 = 4\). That's why we come to soution \(x^2 + 4x -12=(x+6)(x-2)\)

Now are the steps for factoring quadratic equations \(x^2+bx+c \; \; (a=1)\) by factorization:
    1. Factor \(c\) through any multiple of two integer numbers. We have \(mn=c\)
    2. Find the pair that has the sum equal to \(\frac{b}{a}\). We have \( m+n=b\)
    3. The quadratic equations will be factored as \((x+m)(x+n)\)

Now, let's try with some questions.

Question 3. \(x^2-5x+6=0\)
Question 4. \(x^2-7x+6=0\)
Question 5. \(x^2-2x-8=0\)

This approach is useful if \(a,b,c\) are small, and we could find \(m,n\) easily. However, how could you solve this question?
Question 6. \(5x^2-34x+24=0\)

It's nearly impossible to solve this question under 2 minutes by using 1st approach. We need to use 2nd approach as below.

Approach 2. Factor quadratic equations by making square expression

I'll express the approach by solving Question 6 step by step.

First, note that \((u+v)^2=u^2+2uv+v^2\). We will use this theory to solve quadratic equations

Next, we need to get rig of the coefficient of \(x^2\)

\(5x^2-34x+24 = 5(x^2-\frac{34}{5}x+\frac{24}{5})\). We just need to solve \(x^2-\frac{34}{5}x+\frac{24}{5}\)

Now, we apply the theory above to make square expression:
\(x^2-\frac{34}{5}x+\frac{24}{5}= x^2 - 2 \times x\times \frac{17}{5}+24\)

In this expression, we have \(u=x\) and \(v=\frac{17}{5}\). Hence the 3rd number is \(v^2=\big(\frac{17}{5} \big)^2=\frac{17^2}{5^2}=\frac{289}{25}\)

Hence, we have
\(x^2 - 2 \times x \times \frac{17}{5}+24 \\
=x^2 - 2 \times x\times \frac{17}{5}+\big(\frac{17}{5} \big)^2+\big (24-\frac{289}{25} \big ) \\
=\Big ( x - \frac{17}{5}\Big )^2-\frac{169}{25} \\
= \Big ( x - \frac{17}{5}\Big )^2 - \big ( \frac{13}{5} \big ) ^2 \\
= \Big ( x - \frac{17}{5} - \frac{13}{5} \Big ) \Big ( x - \frac{17}{5} + \frac{13}{5} \Big ) \\
= \Big ( x - \frac{30}{5} \Big ) \Big ( x - \frac{4}{5} \Big ) \\
= (x-6)( x - \frac{4}{5})\)

Now we have \(5x^2-34x+24 = 5(x^2-\frac{34}{5}x+\frac{24}{5}) = 5\Big ( x - 6 \Big ) \Big ( x - \frac{4}{5} \Big ) = (x-6)(5x-4)\)

Now, we could easily solve this equation :-D

Question 7. Solve equation \(x(x+4)=672\)
You may come to this equation if you solve this question
https://gmatclub.com/forum/at-his-regul ... 36642.html

Solution.
\(x(x+4)-672 = x^2+4x - 672 = x^2 + 2 \times x \times 2 + 4 - 676 \\
= (x+2)^2 - 26^2 = (x+2-26)(x+2+26)=(x-24)(x+28)\)

Now, you could practice with these questions

Question 8. \(4x^2-19x+12=0\)
Question 9. \(x^2-13x-130=0\)
Question 10. \(5x^2 - 10x + 6=0\)
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Re: How to solve quadratic equations - Factor quadratic equations  [#permalink]

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New post 25 Jul 2019, 15:08
You literally never need to make a square expression.

For the second example, it has to be (5x - )(x - ) or (5x - )(x + ), because you need to make 5x^2, and 5 only has the factors of 5 and 1.

So now we have to find two numbers that multiply together to 24, and add together to -34 (when one is multiplied by 5).

12 and 2 doesn't work because you either get -60 and -2, or -10 and -12.

8 and 3 doesn't work because you either get -40 and -3, or -15 and -8.

6 and 4 work because you get -30 and -4. So you're done: pair up the 6 with the -5 and the 4 with the - 1.

Ta da! And you don't have to worry about 289/25 as a fraction.
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Re: How to solve quadratic equations - Factor quadratic equations   [#permalink] 25 Jul 2019, 15:08
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