Show that the statement, p: if a is a real number such that a3 + 4a =0, then a is 0″, is true by direct method?

Sure, let's approach this problem step by step. First, let's recall the statement: p:Ifaisarealnumbersuchthata3+4a=0,thena=0p:Ifaisarealnumbersuchthata3+4a=0,thena=0 We want to prove this statement using the direct method, which means we need to start with the assumption that a3+4a=0a3+4a=0 and then deduce that a=0a=0. Here's the proof: Proof: Assume a3+4a=0a3+4a=0 for some real number aa. Now, let's factor out aa from the equation: a(a2+4)=0a(a2+4)=0 Since aa is a real number, either a=0a=0 or a2+4=0a2+4=0. If a=0a=0, then the statement a=0a=0 holds true. If a2+4=0a2+4=0, then a2=−4a2=−4. However, there are no real numbers whose square is -4. Thus, this case is not possible. Since both cases lead to a=0a=0, we have shown that if a3+4a=0a3+4a=0, then a=0a=0. Therefore, the statement pp is true by direct method. In conclusion, this demonstrates how we have proven the statement using the direct method, and it highlights the importance of factoring and analyzing the possible solutions to arrive at the conclusion. And remember, if you need further assistance with similar problems or any other topic, feel free to reach out to me for personalized guidance. Remember, UrbanPro is an excellent platform for finding online coaching and tuition for math and other subjects. read less