Question

in simplest radical form, what are the solutions to the quadratic equation 6 = x² - 10x?
quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\\(\bigcirc\\) $x = 5 \pm \sqrt{11}$
\\(\bigcirc\\) $x = 5 \pm \sqrt{19}$
\\(\bigcirc\\) $x = 5 \pm 2\sqrt{19}$
\\(\bigcirc\\) $x = 5 \pm 2\sqrt{31}$

Explanation:

Step1: Rewrite the equation

Rewrite the equation \(6 = x^2 - 10x\) in standard quadratic form \(ax^2+bx+c = 0\). Subtract 6 from both sides: \(x^2-10x - 6=0\). Here, \(a = 1\), \(b=- 10\), \(c=-6\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b=-10\), \(c = - 6\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-10)^{2}-4\times1\times(-6)=100 + 24=124\)? Wait, no, wait, let's recalculate. Wait, the original equation after rearrangement: \(x^{2}-10x - 6=0\)? Wait, no, wait the original equation is \(6=x^{2}-10x\), so moving 6 to the other side: \(x^{2}-10x-6 = 0\)? Wait, no, wait, maybe I made a mistake. Wait, let's check again. Wait, the equation is \(6=x^{2}-10x\), so \(x^{2}-10x - 6=0\)? Wait, no, wait, maybe the user made a typo? Wait, no, looking at the options, the discriminant should be such that when we complete the square or use quadratic formula, let's try completing the square. Let's rewrite \(x^{2}-10x=6\). Completing the square: \(x^{2}-10x + 25=6 + 25\) (since \((\frac{-10}{2})^{2}=25\)). So \((x - 5)^{2}=31\). Then \(x-5=\pm\sqrt{31}\), so \(x = 5\pm\sqrt{31}\)? Wait, no, the options have \(5\pm2\sqrt{31}\)? Wait, no, wait, maybe I misread the original equation. Wait, the original equation is \(6=x^{2}-10x\)? Wait, no, maybe it's \(6 = x^{2}-10x\) or maybe \(6=x^{2}-10x\) is wrong? Wait, no, let's check the quadratic formula again. Wait, if the equation is \(x^{2}-10x-6=0\), then \(a = 1\), \(b=-10\), \(c=-6\). Then discriminant \(b^{2}-4ac=(-10)^{2}-4\times1\times(-6)=100 + 24 = 124\), which is not a perfect square. But the options have \(\sqrt{31}\) or \(\sqrt{19}\). Wait, maybe the original equation is \(6=x^{2}-10x\) is incorrect, maybe it's \(6 = x^{2}-10x\) or maybe \(6=x^{2}-10x\) is a typo, and it's \(6=x^{2}-10x\) or maybe \(6=x^{2}-10x\) should be \(6=x^{2}-10x\) but let's check the options. Wait, the options are \(x = 5\pm\sqrt{31}\), \(x = 5\pm\sqrt{19}\), \(x = 5\pm2\sqrt{19}\), \(x = 5\pm2\sqrt{31}\). Wait, maybe the original equation is \(6 = x^{2}-10x\) is wrong, and it's \(6 = x^{2}-10x\) or maybe \(6=x^{2}-10x\) is \(6=x^{2}-10x\) and when we use quadratic formula, \(a = 1\), \(b=-10\), \(c=-6\), discriminant is 124, which is \(4\times31\), so \(\sqrt{124}=\sqrt{4\times31}=2\sqrt{31}\). Ah! There we go. So \(x=\frac{10\pm\sqrt{124}}{2}=\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\)? Wait, no, \(\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\)? Wait, no, \(2\sqrt{31}\) divided by 2 is \(\sqrt{31}\), so \(x=\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\)? But the last option is \(x = 5\pm2\sqrt{31}\). Wait, I must have made a mistake. Wait, let's re-express the equation. Let's start over.

Given the equation \(6=x^{2}-10x\), rewrite it as \(x^{2}-10x - 6=0\).

Quadratic formula: \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 1\), \(b=-10\), \(c=-6\).

So, \(x=\frac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(-6)}}{2(1)}=\frac{10\pm\sqrt{100 + 24}}{2}=\frac{10\pm\sqrt{124}}{2}\).

Simplify \(\sqrt{124}\): \(\sqrt{4\times31}=2\sqrt{31}\). So, \(x=\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\)? Wait, no, \(\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\) (dividing numerator and denominator by 2). But the last option is \(x = 5\pm2\sqrt{31}\). Wait, this is a contradiction. Wait, maybe the original equation is \(6=x^{2}-10x\) is wrong, and it's \(6=x^{2}-10x\) or maybe \(6=x^{2}-10x\) should be \(6=x^{2}-10x\) but let's check the options again. Wait, the first option is \(x = 5\pm\sqrt{31}\), which would be the case if the equation is \(x^{2}-10x - 6=0\) and we simplify \(\frac{10\pm2\sqrt{31}}{2}\) to \(5\pm\sqrt{31}\). So maybe that's the intended solution. So the correct answer is \(x = 5\pm\sqrt{31}\), which is the first option? Wait, no, the first option is \(x = 5\pm\sqrt{31}\), the last option is \(x = 5\pm2\sqrt{31}\). Wait, I think I made a mistake in the discriminant. Wait, if the equation is \(x^{2}-10x-6=0\), then \(b^{2}-4ac=100 + 24 = 124=4\times31\), so \(\sqrt{124}=2\sqrt{31}\), so \(x=\frac{10\pm2\sqrt{31}}{2}=5\pm\sqrt{31}\). So the first option is \(x = 5\pm\sqrt{31}\), which would be correct.

Wait, but let's check the completing the square method again. \(x^{2}-10x=6\). Add 25 to both sides: \(x^{2}-10x + 25=6 + 25\), so \((x - 5)^{2}=31\), so \(x - 5=\pm\sqrt{31}\), so \(x = 5\pm\sqrt{31}\). Yes, that's correct. So the first option is \(x = 5\pm\sqrt{31}\), so the answer is the first option.

Answer:

\(x = 5\pm\sqrt{31}\) (the first option: \(x = 5\pm\sqrt{31}\))