How can I find the solutions to this quadratic inequality?
Separate x, like that:x^2-8x=x*(x-8)now you find the zeroes of the function (the solutions that will give zero to this function):so you are in a situation of:a*b=0and: a=xb=(x-8)when a*b is zero? when a=0 OR b=0it means:x=0 OR x-8=0 so we have two solutions that gives us zero to the funtion:x=0 OR x=8try, just use the function (x^2-8*x) and test if this is right.now you can compose a draft, a very simple one. You know that if:a<0 AND b>0 then a*b is negative a >0 AND b>0 then a*b is positivea<0 AND b<0 then a*b is positivea>0 AND b<0 then a*B is negativeyou can do this grafically like that:when a is positive?a is just x, so i positive when x is positive--------------------o++++++++++++++++++ 0as you can see up here in this line we see that before zero, a is negative, after zero a is positive. for x=0 then a is just zero (the big dot between - and + ).when b is positive? b=x-8, or: x-8>0 is x>8, lets draw this:-------------------------------------------o++++++++++++++++++++++ 8in 8 b is just zero.Now lets draw them together:--------------------o++++++++++++++++++ 0-------------------------------------------o++++++++++++++++++++++ 8and we know that when signs are the same a*b will be positive, otherwise negative. So we can look for it, drawing vertical lines:--------------------|o++++++++++++++++++ |0--------------------|-----------------------|o++++++++++++++++++++++ | |8+++++++++++ |-----------------------|++++++++++++++++++++++and the third orizontal line should be a*b. Let's check the result:x^2-8*x>0 - Wolfram|Alphait seems ok! But now we want also the solution for the equality with zero, you can see it in the first question, it was: x^2-8x>=0it means that you have to include the dotted numbers in the final solution, that is:x<=0 AND x>=8or using intervals: (-inf,0] [8,+inf)I hope is ok even if it is written very fast ;)
An inequality ____ has a real number solution. a.always b.sometime c.never?
Sometimes. For example, x > 3 has infinite solutions. x^2 < -5 has no real solutions.
How can I find solution to this quadratic inequality? Please someone help me.
First one might approach it with trial and error. Is 10 bigger than 24 after you square it and multiply by negative 2? (-200 > 24) No. x=0? No. Neither would negative x satisfy it. So, you might be done now, or get a feel that x squared must be negative, which smells of complex numbers. Now, I was going to suggest manipulating the equation, and checking against the intuitive deduction (being careful about things like what happens to the > sign if you divide by a negative number). However, order operations > and < aren't defined on complex numbers. So rather than letting x = a+bi, you should just prove that there is no such x.Suppose -2x^2 > 24Then x^2 < -12But that's wrong, as x^2 is nonnegative for x in R, qed.Note that "take the square root of both sides" of an inequality is subtly wrong.
Prove that every cubic inequality has infinitely many real solutions?
A simple answer is to state that a cubic inequality results in a domain of all real numbers. Plugging in any of these numbers will lead to a range of all real numbers. Thus, any inequality can be satisfied as the range of any cubic inequality is all real numbers.
Write an inequality statement whose solution is some but not all real numbers.?
That means use < or >. So something like 4x > 8 which is true for numbers > 2 but not true for numbers ≤ 2.
Solve the inequality in terms of intervals and illustrate the solution set on a real number line?
2x+7>3 2x>-4 x > -2 You would put a open circle at -2 and then draw a solid line extending to the right going to infinity. The interval would be (-2, infinity) or -2< x
Algebra :: Problems with all real numbers and/or no solution?
I have taken the liberty of re-ordering and enhancing the wording of the questions you are asking: ➊ Can you describe what it means to have a solution of all real numbers? Perhaps a simple example will help. Here is an equation that has as its solution ALL the real numbers: 3x + 4x = 7x No matter what number you replace x with, you will ALWAYS get a true statement. If you were to simplify the equation, you would end up with .... 0 = 0, which is true, but the variable x has apparently "disappeared" into proverbial "thin air". This is part of the answer to your 3rd Q: "How can I recognize an 'all real numbers' solution?" ... .... when an equation reduces/simplifies to 0 = 0 (An inequality similar to this would be 3x + 4x ≥ 7x ➞ 0 ≥ 0, which also is ALWAYS true) ➋ Can you describe what it means to have no real solution? How about this equation: 2x + 3 = 2x + 4 No matter what number you replace x with, you will NEVER get a true statement. If you were to simplify the equation, you would end up with .... 0 = 1, which is obviously FALSE, and once again the variable x has apparently "disappeared" into proverbial "thin air". This is the other part of the answer to your 3rd Q: "How can I recognize an 'no real numbers' solution?" ... ... when an equation reduces/simplifies to a = b (where a ≠ b), and is obviously false. (An inequality similar to this would be 3x + 4 ≥ 3x + 5 ➞ 4 ≥ 5 ➞ 0 ≥ 1, which also is obviously false.) How about this equation: x² = – 4, Can you think of ANY real number that will make this equation true? Neither can I! That's because there IS NO REAL NUMBER that will do the job. There IS however a type of number that WILL help here ... they are called the IMAGINARY numbers (no I am not kidding you!), a.k.a. The Complex Numbers. ➌ How can I recognize all real numbers or no solution in the problem? I hope I have have answered this along the way. Hope this helps! Cheers! ☺ .
How do you graph an inequality with all real numbers?
EDITED after additional details. You can ignore that stuff below for now, but keep it for later. You'll need it. The "graph" would be a solid line along your x-axis with arrows on each side to indicate it continues infinitely in both directions. Example: <=========> -2 -1 0 1 2 3 The interval notation would be as follows: (-∞, ∞) {- infinity to infinity} Since infinity, the number, is undefined, you write the interval like this: (-∞, ∞) instead of [-∞, ∞] (good to know). ---------------------------------------... For y = x +1, you would draw a solid line through points (0, 1) (1, 2) (2, 3) and place arrows on each side <-------> to indicate that the line goes infinitely in both directions. For y > x +1, the line is no longer part of the solution. It is the boundary. To indicate this, you use dashes <- - - - > instead of a solid line. Since y in (0, 1) (1, 2) (2, 3) is > 1, 2 & 3, you shade the area ABOVE your line to indicate that this area is your solution set. For y < x +1, since y in (0, 1) (1, 2) (2, 3) is < 1, 2 & 3, you would shade the area BELOW your line to indicate that this area is your solution set.
How can equations and inequalities be used in real life situations?
You basically used equations everyday without even notice. Like when you have to calculate “How many dollars does the sum of your grocery stuffs equal to?”. Or a more complicated situation, if you need to be in the meeting at 9:30am, the travel distance is 10km and you can only leave after 8:30am, your driving speed should equal “x” km/h.”Inequalities are used for comparison. “He is taller than me.” or “The iPhone with larger memory capacity is more expensive.”. A more complicated example is also with calculating your driving speed. “Now if there is expected traffic jam for 2km. You get out of the traffic jam at 9:00am, you have to drive faster than “x” km/h to compensate for your loss time.”Just some simple examples. There are a lots more out there, just look around.
Absolute Value Equations & Inequalities as Solution Sets?
For [-8,8]: The easiest way to think of how to write this as an absolute value statement is to find the halfway point between -8 and 8. That is the average of the two numbers. (-8 + 8)/2 = 0. 0 then is the point equally far from -8 and 8. So as an absolute value statement: abs(x - 0) <= 8. This gives the inequality statement: -8 <= x - 0 <= 8; -8 <= x <= 8. In interval notation [-8, 8]. The square brackets indicate that -8 and 8 are included in the solution. For (-4, 6): Again find the point midway between -4 and 6. (-4 + 6)/2 = 1. 6 and -4 must be equally far away from this point. Both are 5 units away. So the absolute value statement reflecting this is abs(x - 1) < 5. The inequality statement would then be -5 < x -1 < 5, and the solution is -4 < x < 6. In interval notation this becomes (-4, 6). Parentheses indicate that -4 and 6 are not included in the solution.