Mathematics
30.12.2022 13:48
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Determine the value of k so that 2x^2+kx+5=0 has each type of solutions: 2 real

Determine the value of k so that 2x^2+kx+5=0 has each type of solutions:
2 real solutions:
2complex,nonreal solutions:
exactly 1 real solution:
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lovelyheart5337
lovelyheart5337
4,9(48 marks)
2x^2+kx+5=0 \\ \\
a=2 \\ b=k \\ c=5 \\ \Delta=b^2-4ac=k^2-4 \times 2 \times 5=k^2-40

When the discriminant Δ is greater than 0, the equation has 2 real solutions:
k^2-40  0 \\
k^240 \\
k\sqrt{40} \ \lor \ k2\sqrt{10} \ \lor \ k

When the discriminant Δ is less than 0, the equation has 2 complex and no real solutions:
k^2-40

When the discriminant Δ is equal to 0, the equation has exactly 1 real solution:
k^2-40=0 \\
k^2=40 \\
k=\sqrt{40} \ \lor \ k=-\sqrt{40} \\
k=2\sqrt{10} \ \lor \ k=-2\sqrt{10} \\
\boxed{k \in \{ -2\sqrt{10}, 2\sqrt{10} \} }
lasvegas5811
lasvegas5811
5,0(13 marks)
11x+.06(10,000-x)=700 .11x+600-.06x=700 .05x=700-600 .05x=100 x=100/.05 x=2,000 the amount invested @ 11%. 10,000-2,000=8,ooo the amount invested @ 6%. Proof: .11*2,000+.06*8,000=700 220+480700 700=700

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