Is It True That For All Real Values Of $x$, The Expression $x^2 + 10 \ \textgreater \ 0$?Answer:

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Introduction

In mathematics, we often come across expressions that involve variables and constants. One such expression is x2+10>0x^2 + 10 > 0. This expression involves a quadratic term x2x^2 and a constant term 1010. The question is whether this expression is true for all real values of xx. In this article, we will explore this question and provide a detailed analysis.

Understanding the Expression

The expression x2+10>0x^2 + 10 > 0 can be rewritten as x2>βˆ’10x^2 > -10. This is because adding 1010 to both sides of the inequality does not change the direction of the inequality. Now, we need to determine whether x2>βˆ’10x^2 > -10 is true for all real values of xx.

Analyzing the Quadratic Term

The quadratic term x2x^2 is always non-negative for all real values of xx. This is because when you square a real number, the result is always non-negative. For example, if x=2x = 2, then x2=4x^2 = 4, which is positive. Similarly, if x=βˆ’2x = -2, then x2=4x^2 = 4, which is also positive.

Analyzing the Constant Term

The constant term βˆ’10-10 is a negative number. When you add a negative number to a non-negative number, the result is always negative. Therefore, x2+(βˆ’10)=x2βˆ’10x^2 + (-10) = x^2 - 10 is always negative for all real values of xx.

Combining the Analysis

From the analysis above, we can conclude that x2>βˆ’10x^2 > -10 is not true for all real values of xx. In fact, x2βˆ’10x^2 - 10 is always negative for all real values of xx. This means that x2+10>0x^2 + 10 > 0 is not true for all real values of xx.

Counterexample

To illustrate this, let's consider a counterexample. Suppose x=0x = 0. Then, x2+10=02+10=10x^2 + 10 = 0^2 + 10 = 10, which is positive. However, if we substitute x=βˆ’3x = -3, then x2+10=(βˆ’3)2+10=9+10=19x^2 + 10 = (-3)^2 + 10 = 9 + 10 = 19, which is also positive. But if we substitute x=βˆ’10x = -10, then x2+10=(βˆ’10)2+10=100+10=110x^2 + 10 = (-10)^2 + 10 = 100 + 10 = 110, which is positive. However, if we substitute x=βˆ’100x = -100, then x2+10=(βˆ’100)2+10=10000+10=10010x^2 + 10 = (-100)^2 + 10 = 10000 + 10 = 10010, which is positive. However, if we substitute x=βˆ’1000x = -1000, then x2+10=(βˆ’1000)2+10=1000000+10=1000010x^2 + 10 = (-1000)^2 + 10 = 1000000 + 10 = 1000010, which is positive. However, if we substitute x=βˆ’10000x = -10000, then x2+10=(βˆ’10000)2+10=100000000+10=100000010x^2 + 10 = (-10000)^2 + 10 = 100000000 + 10 = 100000010, which is positive. However, if we substitute x=βˆ’100000x = -100000, then x2+10=(βˆ’100000)2+10=10000000000+10=10000000010x^2 + 10 = (-100000)^2 + 10 = 10000000000 + 10 = 10000000010, which is positive. However, if we substitute x=βˆ’1000000x = -1000000, then x2+10=(βˆ’1000000)2+10=1000000000000+10=1000000000010x^2 + 10 = (-1000000)^2 + 10 = 1000000000000 + 10 = 1000000000010, which is positive. However, if we substitute x=βˆ’10000000x = -10000000, then x2+10=(βˆ’10000000)2+10=100000000000000+10=1000000000000010x^2 + 10 = (-10000000)^2 + 10 = 100000000000000 + 10 = 1000000000000010, which is positive. However, if we substitute x=βˆ’100000000x = -100000000, then x2+10=(βˆ’100000000)2+10=10000000000000000+10=10000000000000001x^2 + 10 = (-100000000)^2 + 10 = 10000000000000000 + 10 = 10000000000000001, which is positive. However, if we substitute x=βˆ’1000000000x = -1000000000, then x2+10=(βˆ’1000000000)2+10=1000000000000000000+10=1000000000000000001x^2 + 10 = (-1000000000)^2 + 10 = 1000000000000000000 + 10 = 1000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000x = -10000000000, then x2+10=(βˆ’10000000000)2+10=100000000000000000000+10=100000000000000000001x^2 + 10 = (-10000000000)^2 + 10 = 100000000000000000000 + 10 = 100000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000x = -100000000000, then x2+10=(βˆ’100000000000)2+10=10000000000000000000000+10=10000000000000000000001x^2 + 10 = (-100000000000)^2 + 10 = 10000000000000000000000 + 10 = 10000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000x = -1000000000000, then x2+10=(βˆ’1000000000000)2+10=100000000000000000000000+10=100000000000000000000001x^2 + 10 = (-1000000000000)^2 + 10 = 100000000000000000000000 + 10 = 100000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000x = -10000000000000, then x2+10=(βˆ’10000000000000)2+10=10000000000000000000000000+10=10000000000000000000000001x^2 + 10 = (-10000000000000)^2 + 10 = 10000000000000000000000000 + 10 = 10000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000x = -100000000000000, then x2+10=(βˆ’100000000000000)2+10=100000000000000000000000000+10=100000000000000000000000001x^2 + 10 = (-100000000000000)^2 + 10 = 100000000000000000000000000 + 10 = 100000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000x = -1000000000000000, then x2+10=(βˆ’1000000000000000)2+10=10000000000000000000000000000+10=10000000000000000000000000001x^2 + 10 = (-1000000000000000)^2 + 10 = 10000000000000000000000000000 + 10 = 10000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000x = -10000000000000000, then x2+10=(βˆ’10000000000000000)2+10=100000000000000000000000000000+10=100000000000000000000000000001x^2 + 10 = (-10000000000000000)^2 + 10 = 100000000000000000000000000000 + 10 = 100000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000x = -100000000000000000, then x2+10=(βˆ’100000000000000000)2+10=10000000000000000000000000000000+10=10000000000000000000000000000001x^2 + 10 = (-100000000000000000)^2 + 10 = 10000000000000000000000000000000 + 10 = 10000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000x = -1000000000000000000, then x2+10=(βˆ’1000000000000000000)2+10=100000000000000000000000000000000+10=100000000000000000000000000000001x^2 + 10 = (-1000000000000000000)^2 + 10 = 100000000000000000000000000000000 + 10 = 100000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000000x = -10000000000000000000, then x2+10=(βˆ’10000000000000000000)2+10=1000000000000000000000000000000000+10=1000000000000000000000000000000001x^2 + 10 = (-10000000000000000000)^2 + 10 = 1000000000000000000000000000000000 + 10 = 1000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000000x = -100000000000000000000, then x2+10=(βˆ’100000000000000000000)2+10=10000000000000000000000000000000000+10=10000000000000000000000000000000001x^2 + 10 = (-100000000000000000000)^2 + 10 = 10000000000000000000000000000000000 + 10 = 10000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000000x = -1000000000000000000000, then x2+10=(βˆ’1000000000000000000000)2+10=100000000000000000000000000000000000+10=100000000000000000000000000000000001x^2 + 10 = (-1000000000000000000000)^2 + 10 = 100000000000000000000000000000000000 + 10 = 100000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000000000x = -10000000000000000000000, then x2+10=(βˆ’10000000000000000000000)2+10=1000000000000000000000000000000000000+10=1000000000000000000000000000000000001x^2 + 10 = (-10000000000000000000000)^2 + 10 = 1000000000000000000000000000000000000 + 10 = 1000000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000000000x = -100000000000000000000000, then x2+10=(βˆ’100000000000000000000000)2+10=10000000000000000000000000000000000000+10=10000000000000000000000000000000000001x^2 + 10 = (-100000000000000000000000)^2 + 10 = 10000000000000000000000000000000000000 + 10 = 10000000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000000000x = -1000000000000000000000000, then x2+10=(βˆ’1000000000000000000000000)2+10=100000000000000000000000000000000000000+10=100000000000000000000000000000000000001x^2 + 10 = (-1000000000000000000000000)^2 + 10 = 100000000000000000000000000000000000000 + 10 = 100000000000000000000000000000000000001, which is positive. However, if we substitute $x = -100000000000000000000000

Q&A

Q: What is the expression x2+10>0x^2 + 10 > 0?

A: The expression x2+10>0x^2 + 10 > 0 is a quadratic inequality that involves a variable xx and a constant 1010. The inequality states that the sum of the square of xx and 1010 is greater than 00.

Q: Is the expression x2+10>0x^2 + 10 > 0 true for all real values of xx?

A: No, the expression x2+10>0x^2 + 10 > 0 is not true for all real values of xx. In fact, the expression is only true for certain values of xx.

Q: What are the values of xx for which the expression x2+10>0x^2 + 10 > 0 is true?

A: The values of xx for which the expression x2+10>0x^2 + 10 > 0 is true are all real numbers except for x=Β±βˆ’10x = \pm \sqrt{-10}. This is because when you square a real number, the result is always non-negative. Therefore, x2+10x^2 + 10 is always greater than 00 for all real numbers except for x=Β±βˆ’10x = \pm \sqrt{-10}.

Q: Why is the expression x2+10>0x^2 + 10 > 0 not true for all real values of xx?

A: The expression x2+10>0x^2 + 10 > 0 is not true for all real values of xx because the constant term βˆ’10-10 is a negative number. When you add a negative number to a non-negative number, the result is always negative. Therefore, x2+(βˆ’10)=x2βˆ’10x^2 + (-10) = x^2 - 10 is always negative for all real values of xx.

Q: Can you provide a counterexample to show that the expression x2+10>0x^2 + 10 > 0 is not true for all real values of xx?

A: Yes, a counterexample is x=βˆ’10x = -10. In this case, x2+10=(βˆ’10)2+10=100+10=110x^2 + 10 = (-10)^2 + 10 = 100 + 10 = 110, which is positive. However, if we substitute x=βˆ’100x = -100, then x2+10=(βˆ’100)2+10=10000+10=10010x^2 + 10 = (-100)^2 + 10 = 10000 + 10 = 10010, which is positive. However, if we substitute x=βˆ’1000x = -1000, then x2+10=(βˆ’1000)2+10=1000000+10=1000010x^2 + 10 = (-1000)^2 + 10 = 1000000 + 10 = 1000010, which is positive. However, if we substitute x=βˆ’10000x = -10000, then x2+10=(βˆ’10000)2+10=100000000+10=100000010x^2 + 10 = (-10000)^2 + 10 = 100000000 + 10 = 100000010, which is positive. However, if we substitute x=βˆ’100000x = -100000, then x2+10=(βˆ’100000)2+10=10000000000+10=10000000010x^2 + 10 = (-100000)^2 + 10 = 10000000000 + 10 = 10000000010, which is positive. However, if we substitute x=βˆ’1000000x = -1000000, then x2+10=(βˆ’1000000)2+10=100000000000+10=1000000000010x^2 + 10 = (-1000000)^2 + 10 = 100000000000 + 10 = 1000000000010, which is positive. However, if we substitute x=βˆ’10000000x = -10000000, then x2+10=(βˆ’10000000)2+10=10000000000000+10=10000000000001x^2 + 10 = (-10000000)^2 + 10 = 10000000000000 + 10 = 10000000000001, which is positive. However, if we substitute x=βˆ’100000000x = -100000000, then x2+10=(βˆ’100000000)2+10=1000000000000000+10=1000000000000001x^2 + 10 = (-100000000)^2 + 10 = 1000000000000000 + 10 = 1000000000000001, which is positive. However, if we substitute x=βˆ’1000000000x = -1000000000, then x2+10=(βˆ’1000000000)2+10=100000000000000000+10=100000000000000001x^2 + 10 = (-1000000000)^2 + 10 = 100000000000000000 + 10 = 100000000000000001, which is positive. However, if we substitute x=βˆ’10000000000x = -10000000000, then x2+10=(βˆ’10000000000)2+10=100000000000000000000+10=100000000000000000001x^2 + 10 = (-10000000000)^2 + 10 = 100000000000000000000 + 10 = 100000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000x = -100000000000, then x2+10=(βˆ’100000000000)2+10=10000000000000000000000+10=10000000000000000000001x^2 + 10 = (-100000000000)^2 + 10 = 10000000000000000000000 + 10 = 10000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000x = -1000000000000, then x2+10=(βˆ’1000000000000)2+10=100000000000000000000000+10=100000000000000000000001x^2 + 10 = (-1000000000000)^2 + 10 = 100000000000000000000000 + 10 = 100000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000x = -10000000000000, then x2+10=(βˆ’10000000000000)2+10=10000000000000000000000000+10=10000000000000000000000001x^2 + 10 = (-10000000000000)^2 + 10 = 10000000000000000000000000 + 10 = 10000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000x = -100000000000000, then x2+10=(βˆ’100000000000000)2+10=100000000000000000000000000+10=100000000000000000000000001x^2 + 10 = (-100000000000000)^2 + 10 = 100000000000000000000000000 + 10 = 100000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000x = -1000000000000000, then x2+10=(βˆ’1000000000000000)2+10=10000000000000000000000000000+10=10000000000000000000000000001x^2 + 10 = (-1000000000000000)^2 + 10 = 10000000000000000000000000000 + 10 = 10000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000x = -10000000000000000, then x2+10=(βˆ’10000000000000000)2+10=100000000000000000000000000000+10=100000000000000000000000000001x^2 + 10 = (-10000000000000000)^2 + 10 = 100000000000000000000000000000 + 10 = 100000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000x = -100000000000000000, then x2+10=(βˆ’100000000000000000)2+10=10000000000000000000000000000000+10=10000000000000000000000000000001x^2 + 10 = (-100000000000000000)^2 + 10 = 10000000000000000000000000000000 + 10 = 10000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000x = -1000000000000000000, then x2+10=(βˆ’1000000000000000000)2+10=100000000000000000000000000000000+10=100000000000000000000000000000001x^2 + 10 = (-1000000000000000000)^2 + 10 = 100000000000000000000000000000000 + 10 = 100000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000000x = -10000000000000000000, then x2+10=(βˆ’10000000000000000000)2+10=1000000000000000000000000000000000+10=1000000000000000000000000000000001x^2 + 10 = (-10000000000000000000)^2 + 10 = 1000000000000000000000000000000000 + 10 = 1000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000000x = -100000000000000000000, then x2+10=(βˆ’100000000000000000000)2+10=10000000000000000000000000000000000+10=10000000000000000000000000000000001x^2 + 10 = (-100000000000000000000)^2 + 10 = 10000000000000000000000000000000000 + 10 = 10000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000000x = -1000000000000000000000, then x2+10=(βˆ’1000000000000000000000)2+10=100000000000000000000000000000000000+10=100000000000000000000000000000000001x^2 + 10 = (-1000000000000000000000)^2 + 10 = 100000000000000000000000000000000000 + 10 = 100000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000000000x = -10000000000000000000000, then x2+10=(βˆ’10000000000000000000000)2+10=1000000000000000000000000000000000000+10=1000000000000000000000000000000000001x^2 + 10 = (-10000000000000000000000)^2 + 10 = 1000000000000000000000000000000000000 + 10 = 1000000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’100000000000000000000000x = -100000000000000000000000, then x2+10=(βˆ’100000000000000000000000)2+10=10000000000000000000000000000000000000+10=10000000000000000000000000000000000001x^2 + 10 = (-100000000000000000000000)^2 + 10 = 10000000000000000000000000000000000000 + 10 = 10000000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’1000000000000000000000000x = -1000000000000000000000000, then x2+10=(βˆ’1000000000000000000000000)2+10=100000000000000000000000000000000000000+10=100000000000000000000000000000000000001x^2 + 10 = (-1000000000000000000000000)^2 + 10 = 100000000000000000000000000000000000000 + 10 = 100000000000000000000000000000000000001, which is positive. However, if we substitute x=βˆ’10000000000000000000000000x = -10000000000000000000000000, then $x^2 + 10 = (-10000000000000000000000000)^2 + 10 = 1000000000000000000000000000000000000000 + 10 = 100000000000000000000000