Any value in the interval (-∞,-2√5] ∪[2√5,∞) will cause the quadratic equation x2+bx+5 = 0 to have two real number solutions.
Given quadratic equation is:
[tex]x^{2} +bx+5=0[/tex]
Any equation of the form [tex]ax^{2} +bx+c=0[/tex] is called a quadratic equation where a≠0.
To have two real number solutions the discriminant of a quadratic equation should be greater than or equal to zero.
[tex]D\geq 0[/tex]
[tex]b^{2} -4(1)(5)\geq 0[/tex]
[tex]b^{2}-20 \geq 0[/tex]
[tex]b^{2} -(2\sqrt{5}) ^{2}\geq 0[/tex]
[tex](b+2\sqrt{5} )(b-2\sqrt{5} )\geq 0[/tex]
b∈[tex](-\infty,-2\sqrt{5}][/tex]∪[tex][2\sqrt{5},\infty)[/tex]
Hence, any value in the interval (-∞,-2√5] ∪[2√5,∞) will cause the quadratic equation x2+bx+5 = 0 to have two real number solutions.
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