Question

Which equation has the solutions x = startfraction 5 plus-or-minus 2 startroot 7 endroot over 3 endfraction?

Answer 1

The equation that has the solution $x = \frac{5 \pm \sqrt{7}}{3}$ is 3x^2 - 10x + 6 = 0

How to determine the equation?

The solution is given as:

$x = \frac{5 \pm \sqrt{7}}{3}$

The solution to a quadratic equation is

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

By comparing both equations, we have:

-b = 5

b^2 - 4ac = 7

2a = 3

Solve for b in -b = 5

b = -5

Solve for a in 2a = 3

a = 1.5

Substitute values for a and b in b^2 - 4ac = 7

(-5)^2 - 4 * 1.5c = 7

Evaluate

25 - 6c = 7

Subtract 25 from both sides

-6c = -18

Divide by - 6

c = 3

So, we have:

a = 1.5

b =  -5

c = 3

A quadratic equation is represented as:

ax^2 + bx + c = 0

So, we have:

1.5x^2 - 5x +3 = 0

Multiply through by 2

3x^2 - 10x + 6 = 0

Hence, the equation that has the solution $x = \frac{5 \pm \sqrt{7}}{3}$ is 3x^2 - 10x + 6 = 0

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