Question

A quadratic equation of the form 0 = ax2 + bx + c has one real number solution. Which could be the equation? 0 = 2x2 – 4x + 1 0 = 2x2 – 5x + 3 0 = –2x2 – 4x – 2 0 = –2x2 – 3x – 1

Answer 1

answer is c

Explanation:

Answer 2

`0 = -2x^2 - 4x - 2`

Explanation:

To solve this problem we must calculate the discriminant of each quadratic function.

Let
`ax ^ 2 + bx + c` be a quadratic function with a, b and c their real coefficients, then:

If
`b ^ 2 -4ac> 0` the function has 2 real solutions.

If
`b ^ 2 -4ac = 0` the function has 1 double real solution

If
`b ^ 2 -4ac <0` the function has complex solutions.

Now we calculate the discriminant for each of the functions:

`0 = 2x^2 - 4x + 1`

`(-4) ^ 2 -4(2)(1) = 8` (Two real solutions)

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`0 = 2x^2- 5x + 3`

`(-5) ^ 2 -4(2)(3) = 1` (Two real solutions)

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`0 = -2x^2 - 4x - 2`

`(-4) ^ 2 -4(-2)(- 2) = 0` (A real double solution)

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`0 = -2x^2 - 3x - 1`

`(-3) ^ 2 -4(-2)(- 1) = 1` (Two real solutions)

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The equation we want is:

`0 = -2x^2 - 4x - 2`