Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions? –5 –3 3 4

b = -5

Step-by-step explanation:

You needy to calculate the Discriminant = b² - 4ac

If D is zero there is one real solution,

If D less then zero there are no real solutions

Only if D is bigger then zero, there are two real solutions.

x² - 5x + 5 D > 0 so 2 real solutions

see attachment.

x² - 3x + 5 D < 0 so no real solutions

x² + 3x + 5 D < 0 so no real solutions

x² + 4x + 5 D < 0 so no real solutions

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

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