GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 05 Dec 2019, 22:16 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Author Message
TAGS:

### Hide Tags

Retired Moderator V
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1323
Location: Viet Nam

### Show Tags

2
8

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. $$5x^2 - 10x + 6=0$$
_________________
Manager  B
Joined: 20 Jul 2016
Posts: 84

### Show Tags

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.
_________________
Boston-based GMAT, GRE, LSAT Tutor. 750 on the GMAT, 170V/166Q on GRE, 5 stars on Yelp, 5 stars on Google. Always happy to help!

Also, creator of an error log app, 21st Night: https://get21stnight.com/. Use it to answer the question: "What do I study next?" Visit the link to see what I mean. Display posts from previous: Sort by  