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Sagot :
Answer:
k = -1
Step-by-step explanation:
The limit will exist if:
One of the roots of the equation in the numerator is 5. This happens because if this happens, we can simplify with the denominator. So
Solving a quadratic equation:
In the following format:
ax² + bx + c = 0
The solution is given by:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]In this question:
x² + kx - 20 = 0
The solution is:
[tex]x=\frac{-k\pm\sqrt[]{k^2-4\ast1\ast(-20)}}{2}=\frac{-k\pm\sqrt[]{k^2-80}}{2}[/tex]Since we want x = 5.
[tex]\frac{-k+\sqrt[]{k^2+80}}{2}=5[/tex][tex]-k+\sqrt[]{k^2-80}=10[/tex][tex]\sqrt[]{k^2+80}=10+k[/tex][tex](\sqrt[]{k^2+80})^2=(10+k)^2[/tex][tex]k^2+80=100+20k+k^2[/tex][tex]k^2+80-100-20k-k^2=0[/tex][tex]-20-20k=0[/tex][tex]20k=-20[/tex][tex]k=\frac{-20}{20}=-1[/tex]k = -1
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