Question

Solve. Show your work.

5. x^2 − 1x + 19 = −5
6. 7x^2 + x = 0
7. 7????^2 − 14???? = −7
8. 2d^2 + 5d − 12 = 0

Answer 1

5)
`x_1 = (1+ √(95))/(2)`

`x_2 = (1- √(95))/(2)`

6)
`x_1=0`

`x_2 = -(1)/(7)`

7) So for this case we have just one solution
`X=7`

8)
`d_1 = (-5+√(121))/(4)=(3)/(2)`

`d_2 = (-5-√(121))/(4)= -4`

Explanation:

5. x^2 − 1x + 19 = −5

For this case we can rewrite the expression like this:

`x^2 - x + 24=0`

And then we can use the quadratic formula given by:

`x= (-b \pm √(b^2 -4ac))/(2a)`

And for this case a= 1, b = -1 , c = 24, and replacing we got:

`x= (-(-1) \pm √((-1)^2 -4(1)(24)))/(2(1))`

`x_1 = (1+ √(95))/(2)`

`x_2 = (1- √(95))/(2)`

6. 7x^2 + x = 0

For this case we can take common factor first like this:

[tex x (7x +1) = 0[/tex]

So then
`x=0` or
`7x+1=0` and we got that:

`x = -(1)/(7)`

7. 7x^2 − 14x = −7

We can rewrite the expression like this:

`7x^2 -14x +7 = 0`

And then we can use the quadratic formula given by:

`x= (-b \pm √(b^2 -4ac))/(2a)`

And for this case a= 7, b = -14 , c = 7, and replacing we got:

`x= (-(-14) \pm √((-14)^2 -4(7)(7)))/(2(7))`

`x_1 = (14+0)/(14)=1`

`x_2 = (14-0)/(14)= 1`

So for this case we have just one solution
`X=7`

8. 2d^2 + 5d − 12 = 0

We can use the quadratic formula given by:

`d= (-b \pm √(b^2 -4ac))/(2a)`

And for this case a= 2, b = 5 , c = -12, and replacing we got:

`d= (-(5) \pm √((5)^2 -4(2)(-12)))/(2(2))`

`d_1 = (-5+√(121))/(4)=(3)/(2)`

`d_2 = (-5-√(121))/(4)= -4`