Question

Use the quadratic formula to solve each equation.

1. x^2 − 2x = 12 → a = 1, ???? = −2, c = −12 [Watch the negatives.]
2.1 / 2????^2 − 6???? = 2 → a = 1 / 2, ???? = −6, c = −2 [Did you remember the negative?]
3. 2p^2 + 8p = 7 → a = 2, ???? = 8, c = −7
4. 2y^2 + 3y − 5 = 4 → a = 2, ???? = 3, c = −9

Answer 1

1)
`x_1 = (2 - 2√(13))/(2)= 1-√(13)`

`x_2 = (2 + 2√(13))/(2)= 1+√(13)`

2)
`x_1 = (6 - 2√(10))/(1)= 6-2√(10)`

`x_2 = (6 + 2√(10))/(1)= 6+2√(10)`

3)
`p_1 = (-8 - 2√(30))/(4)= -2-(1)/(2)√(30)`

`p_2 = (-8 + 2√(30))/(4)= -2+(1)/(2)√(30)`

4)
`y_1 = (-3 - 9)/(4)=-3`

`y_2 = (-3 + 9)/(4)= (3)/(2)`

Explanation:

The quadratic formula is given by:

`x = (-b \pm √(b^2 -4ac))/(2a)`

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

`x = (-(-2) \pm √((-2)^2 -4(1)(-12)))/(2(1))`

`x_1 = (2 - 2√(13))/(2)= 1-√(13)`

`x_2 = (2 + 2√(13))/(2)= 1+√(13)`

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

`x = (-(-6) \pm √((-6)^2 -4(1/2)(-2)))/(2(1/2))`

`x_1 = (6 - 2√(10))/(1)= 6-2√(10)`

`x_2 = (6 + 2√(10))/(1)= 6+2√(10)`

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

`p = (-(8) \pm √((8)^2 -4(2)(-7)))/(2(2))`

`p_1 = (-8 - 2√(30))/(4)= -2-(1)/(2)√(30)`

`p_2 = (-8 + 2√(30))/(4)= -2+(1)/(2)√(30)`

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

`y = (-(3) \pm √((3)^2 -4(2)(-9)))/(2(2))`

`y_1 = (-3 - 9)/(4)=-3`

`y_2 = (-3 + 9)/(4)= (3)/(2)`