Question

Which parabola will have one real solution with the line y = x – 5? y = x2 + x – 4 y = x2 + 2x – 1 y = x2 + 6x + 9 y = x2 + 7x + 4

Answer 1

Answer:

y = x² + 7x + 4

Step-by-step explanation:

First option

System of equations:

y = x² + x – 4

y = x – 5

Replacing:

x - 5 = x² + x – 4

0 = x² + x – 4 - x + 5

0 = x² + 1

Discriminant:

b²-4*a*c

0²-4*1*1

-4 < 0   then the equation has no real roots

Second option

System of equations:

y = x² + 2x – 1

y = x – 5

Replacing:

x - 5 = x² + 2x – 1

0 = x² + 2x - 1  - x + 5

0 = x² + x + 4

Discriminant:

b²-4*a*c

1²-4*1*4

-15 < 0   then the equation has no real roots

Third option

System of equations:

y = x² + 6x + 9

y = x – 5

Replacing:

x - 5 = x² + 6x + 9

0 = x² + 6x + 9  - x + 5

0 = x² + 5x + 4

Discriminant:

b²-4*a*c

5²-4*1*4

9 > 0 then the equation has two different real roots

Fourth option

System of equations:

y = x² + 7x + 4

y = x – 5

Replacing:

x - 5 = x² + 7x + 4

0 = x² + 7x + 4 - x + 5

0 = x² + 6x + 9

Discriminant:

b²-4*a*c

6²-4*1*9

0  then the equation has one real root