which equation has the solutions $x = \\frac{5\\pm2\\sqrt{7}}{3}$?\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 match the $2a$ in the quadratic - formula. Let's assume $2a = 6$ (so $a = 3$).

Step2: Identify $b$ and $c$

If $a = 3$, then from $-b = 5$, we get $b=-5$. And $b^{2}-4ac = 28$. Substitute $a = 3$ and $b=-5$ into $b^{2}-4ac = 28$. We have $(-5)^{2}-4\times3\times c=28$.

Step3: Solve for $c$

Expand the left - hand side: $25-12c = 28$. Subtract 25 from both sides: $-12c=28 - 25=3$. Then $c=- \frac{1}{4}$. But we can also work in another way. If we start from the general form and cross - multiply the quadratic formula result. If $x=\frac{5\pm2\sqrt{7}}{3}$, then $3x=5\pm2\sqrt{7}$. Rearrange to get $3x - 5=\pm2\sqrt{7}$. Square both sides: $(3x - 5)^{2}=(2\sqrt{7})^{2}$. Expand $(3x - 5)^{2}$ using the formula $(A - B)^{2}=A^{2}-2AB + B^{2}$, where $A = 3x$ and $B = 5$. We have $9x^{2}-30x + 25 = 28$. Rearrange to $9x^{2}-30x-3 = 0$. Divide through by 3 to get $3x^{2}-10x - 1=0$.

Answer:

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