Question

for what values m, -5 is the zeros of polynomial X square -x-(2m -2)​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

For the equation to possess no real roots,

Its discriminant must be <0

=> b^2 - 4ac < 0

=> (m+1)^2 - 4(m)(2m+1) <0

=> m^2+1+2m - 8m^2–4m < 0

=> -7m^2 -2m+1 < 0

Since, the leading coefficient < 0, Therefore the value of the expression would be less than 0 for all values of m, except the ones in between the zeros.

Here,

D = 4 - 4(1)(-7) = 4+28 = 32 = 4 √2,

Therefore, Zeroes =( 2 + 4 √2)/-14 and (2- 4 √2)-14

Hence, The equation would possess no real roots for all real values of m, except those between ( 2 + 4 √2)/-14 and (2- 4 √2)-14

Step-by-step explanation:

For the equation to possess no real roots,

Its discriminant must be <0

=> b^2 - 4ac < 0

=> (m+1)^2 - 4(m)(2m+1) <0

=> m^2+1+2m - 8m^2–4m < 0

=> -7m^2 -2m+1 < 0

Since, the leading coefficient < 0, Therefore the value of the expression would be less than 0 for all values of m, except the ones in between the zeros.

Here,

D = 4 - 4(1)(-7) = 4+28 = 32 = 4 √2,

Therefore, Zeroes =( 2 + 4 √2)/-14 and (2- 4 √2)-14

Hence, The equation would possess no real roots for all real values of m, except those between ( 2 + 4 √2)/-14 and (2- 4 √2)-14