which equation has the solutions $x = \frac{5pm2sqrt{7}}{3}$?\n$3x^{2}-5x + 7 = 0$\n$3x^{2}-5x - 1 =…

Explanation:

Step1: Recall quadratic - formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Given $x=\frac{5\pm2\sqrt{7}}{3}$, we can rewrite it in the form $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, rewrite $\frac{5\pm2\sqrt{7}}{3}$ as $\frac{5\pm\sqrt{28}}{3}$. Then, we want to make the denominator $2a$ and the numerator $-b\pm\sqrt{b^{2}-4ac}$. Let's assume $2a = 3$, then $a=\frac{3}{2}$. But we can also work backward by cross - multiplying to get $3x=5\pm2\sqrt{7}$, then $3x - 5=\pm2\sqrt{7}$. Square both sides: $(3x - 5)^{2}=(2\sqrt{7})^{2}$.

Step2: Expand the left - hand side

Expand $(3x - 5)^{2}$ using the formula $(A - B)^{2}=A^{2}-2AB + B^{2}$, where $A = 3x$ and $B = 5$. So $(3x - 5)^{2}=9x^{2}-30x + 25$. And $(2\sqrt{7})^{2}=28$. Then $9x^{2}-30x + 25=28$.

Step3: Rearrange the equation

Subtract 28 from both sides: $9x^{2}-30x+25 - 28 = 0$, which simplifies to $9x^{2}-30x - 3 = 0$. Divide the entire equation by 3: $3x^{2}-10x - 1 = 0$.

Answer:

$3x^{2}-10x - 1 = 0$ (the fourth option)