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27 May 2017, 08:42
2
6

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

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

Question 1. Solve $$x^2 + 6x +5 = 0$$

Question 2. Solve $$x^2 + 4x -12 =0$$

How did you solve those quetsions? 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

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 $$c=mn$$.

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

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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08 Jul 2018, 11:33
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