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Sagot :
We have the equation:
[tex]49a^2-16=0[/tex]We can factorize this equation as:
[tex]\begin{gathered} 49a^2-16=0 \\ (7a)^2-4^2=0 \\ (7a-4)(7a+4)=0 \end{gathered}[/tex][tex]\begin{gathered} 7a-4=0\longrightarrow a_1=\frac{4}{7} \\ 7a+4=0\longrightarrow a_2=-\frac{4}{7} \end{gathered}[/tex]In this case, we have 2 rational solutions.
If the solution implies the square root of -1, then we would have 2 complex solutions.
If the solution implies a square root that does not have a rational solution, then we have 2 irrational solutions.
We can see it when we apply the quadratic formula:
[tex]x=-\frac{b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a}[/tex]The term with the square root defines what type of solution we have:
If b^2-4ac<0, then we have complex solutions.
If the square root of b^2-4ac does not have a rational solution (b^2-4ac is not a perfect square), then we have irrational solutions.
If b^2-4ac is a perfect square (its square root have a rational solution), we will have rational solutions.
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