Question

Using the quadratic formula to solve 2x^2=4x-7, what's the values of x?

Answer 1

1 + 1.58i , 1 - 1.58i

Explanation:

2x^2=4x-7

2x^2 - 4x + 7 = 0

x = [-(-4) +/- sqrt((-4)^2 - 4 * 2 * 7 )] / 2*2

= [ 4 +/- sqrt (16 - 56)] / 4

= [4 +/- sqrt (-40) ] / 4

= 1 +/- 6.32i / 4

= 1 + 1.58i and 1 - 1.58i (answer).

Answer 2

`\boxed{x = 1 \pm i\sqrt{(5)/( 2)} }\\`

Explanation:

2x² = 4x - 7

2x² - 4x + 7 =0

a = 2; b = -4; c = 7

`y =(-b\pm√(b^2-4ac))/(2a)\\`

`=(4\pm√((-2)^2-4*2*7))/(2*2)\\`

`= 1 \pm(√(16-56))/(4)\\`

`= 1 \pm(√(-40))/(4)\\`

`= 1 \pm(2i√(10))/(4)\\`

`= 1 \pm(i√(10))/(2)\\`

`= 1 \pm i\sqrt{(10)/(4)}\\`

`\boxed{= 1 \pm i\sqrt{(5)/( 2)} }\\`

The graph of y = 2x² - 4x + 7 has a minimum at (1, 5). It never touches the x-axis, so both roots are imaginary.