Question

show what you know quadratic formula
solve: $8n^2 - 4n = 18$
$\frac{1 + i\sqrt{37}}{4}$
$\frac{19}{2}$
$\frac{1 - i\sqrt{37}}{4}$
$\frac{1 - \sqrt{37}}{4}$
$\frac{1 + \sqrt{37}}{4}$
$-9$

Explanation:

Step1: Rewrite in standard form

First, we rewrite the equation \(8n^{2}-4n = 18\) in the standard quadratic form \(ax^{2}+bx + c=0\). Subtract 18 from both sides:
\(8n^{2}-4n - 18=0\)
We can simplify this equation by dividing all terms by 2:
\(4n^{2}-2n - 9 = 0\)
So, \(a = 4\), \(b=- 2\), \(c=-9\)

Step2: Apply the quadratic formula

The quadratic formula is \(n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
Substitute \(a = 4\), \(b=-2\), \(c = - 9\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times4\times(-9)=4 + 144=148\)
Wait, there is a mistake above. Wait, original equation after dividing by 2: \(4n^{2}-2n-9 = 0\), so \(a = 4\), \(b=-2\), \(c=-9\)
\(b^{2}-4ac=(-2)^{2}-4\times4\times(-9)=4 + 144 = 148\)? Wait, no, wait the original equation is \(8n^{2}-4n-18=0\), if we don't divide by 2, \(a = 8\), \(b=-4\), \(c=-18\)
Let's recalculate with \(a = 8\), \(b=-4\), \(c=-18\)
Discriminant \(\Delta=b^{2}-4ac=(-4)^{2}-4\times8\times(-18)=16 + 576=592\)? No, that's not right. Wait, I made a mistake in simplifying. Let's start over.

Original equation: \(8n^{2}-4n=18\)
Subtract 18: \(8n^{2}-4n - 18=0\)
Divide by 2: \(4n^{2}-2n - 9=0\)
Now, \(a = 4\), \(b=-2\), \(c=-9\)
\(b^{2}-4ac=(-2)^{2}-4\times4\times(-9)=4 + 144 = 148\)? Wait, 4*4*9 = 144, and since \(c=-9\), it's \(-4ac=-4\times4\times(-9)=+ 144\), so \(b^{2}-4ac=4 + 144=148\). But \(\sqrt{148}=\sqrt{4\times37}=2\sqrt{37}\)

Now, \(n=\frac{-(-2)\pm\sqrt{148}}{2\times4}=\frac{2\pm2\sqrt{37}}{8}=\frac{1\pm\sqrt{37}}{4}\)

Ah, there we go. So the solutions are \(n=\frac{1+\sqrt{37}}{4}\) and \(n=\frac{1 - \sqrt{37}}{4}\) and also, wait, if the discriminant was negative, we would have imaginary numbers, but in our case, when we calculated with \(a = 4\), \(b=-2\), \(c=-9\), discriminant is \(4 + 144 = 148>0\), so real solutions. Wait, but the options have imaginary numbers. Wait, I must have made a mistake in the sign of \(c\).

Wait, original equation: \(8n^{2}-4n=18\) => \(8n^{2}-4n-18 = 0\). Let's check the options. The options have \(i\) (imaginary unit), which means the discriminant is negative. So I must have messed up the sign of \(c\). Let's re - express the equation as \(8n^{2}-4n-18 = 0\), or maybe I made a mistake in transcribing the equation. Wait, the original equation is \(8n^{2}-4n = 18\), so \(8n^{2}-4n-18=0\). Let's compute the discriminant again: \(b^{2}-4ac=(-4)^{2}-4\times8\times(-18)=16 + 576 = 592\), which is positive. But the options have \(i\), so maybe the original equation is \(8n^{2}-4n=-18\), that is \(8n^{2}-4n + 18=0\)

Ah! That must be the case. If the equation is \(8n^{2}-4n=-18\), then \(8n^{2}-4n + 18=0\)
Now, \(a = 8\), \(b=-4\), \(c = 18\)
Discriminant \(\Delta=b^{2}-4ac=(-4)^{2}-4\times8\times18=16-576=-560\)? No, that's not 37. Wait, maybe the original equation is \(8n^{2}-4n=18\) is wrong, and it's \(8n^{2}-4n=-18\), so \(8n^{2}-4n + 18=0\), divide by 2: \(4n^{2}-2n + 9=0\), then \(a = 4\), \(b=-2\), \(c = 9\)
Discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times4\times9=4 - 144=-140\)? No. Wait, the options have \(\sqrt{37}\), so let's see: if discriminant is \(- 144 + 4=-140\) no. Wait, maybe the equation is \(8n^{2}-4n=18\) is correct, but I made a mistake. Wait, let's check the options. The options have \(1\pm i\sqrt{37}\) over 4, so discriminant is negative, so the equation must be \(8n^{2}-4n=-18\), so \(8n^{2}-4n + 18=0\), divide by 2: \(4n^{2}-2n + 9=0\), \(a = 4\), \(b=-2\), \(c = 9\)
Discriminant \(\Delta=(-2)^{2}-4\times4\times9=4-144=-140\), no. Wait, maybe the original equation is \(8n^{2}-4n=18\) is \(8n^{2}-4n-18 = 0\), and I miscalculated discriminant. Wait, \(8n^{2}-4n-18=0\), \(a = 8\), \(b=-4\), \(c=-18\)
\(b^{2}-4ac=(-4)^{2}-4\times8\times(-18)=16 + 576=592\), \(\sqrt{592}=\sqrt{16\times37}=4\sqrt{37}\)
Then \(n=\frac{-(-4)\pm4\sqrt{37}}{2\times8}=\frac{4\pm4\sqrt{37}}{16}=\frac{1\pm\sqrt{37}}{4}\). But the options have \(1\pm i\sqrt{37}\), so there must be a sign error in \(c\). If the equation is \(8n^{2}-4n=-18\), then \(8n^{2}-4n + 18=0\), \(a = 8\), \(b=-4\), \(c = 18\)
\(b^{2}-4ac=(-4)^{2}-4\times8\times18=16-576=-560\), no. Wait, maybe the equation is \(8n^{2}-4n=18\) is correct, but the options have a typo, or I made a mistake. Wait, the options with \(i\) have discriminant negative, so let's assume that the equation is \(8n^{2}-4n=-18\), so \(8n^{2}-4n + 18=0\), divide by 2: \(4n^{2}-2n + 9=0\), \(a = 4\), \(b=-2\), \(c = 9\)
Discriminant \(\Delta=(-2)^{2}-4\times4\times9=4 - 144=-140\), no. Wait, maybe the original equation is \(8n^{2}-4n=18\) is \(8n^{2}-4n-18 = 0\), and the discriminant is \(592\), but \(\sqrt{592}=4\sqrt{37}\), so \(n=\frac{4\pm4\sqrt{37}}{16}=\frac{1\pm\sqrt{37}}{4}\), which are real solutions. But the options have \(i\) terms, which are complex. So there must be a mistake in the equation. Wait, maybe the equation is \(8n^{2}-4n=-18\), so \(8n^{2}-4n + 18=0\), then \(a = 8\), \(b=-4\), \(c = 18\)
\(b^{2}-4ac=(-4)^{2}-4\times8\times18=16-576=-560\), \(\sqrt{- 560}=\sqrt{16\times(-35)} = 4i\sqrt{35}\), not 37. Hmm.

Wait, let's check the options again. The options are \(\frac{1\pm i\sqrt{37}}{4}\), \(\frac{19}{2}\), \(\frac{1\pm\sqrt{37}}{4}\), \(-9\)

Let's assume that the equation is \(4n^{2}-2n + 9=0\) (divided by 2 from \(8n^{2}-4n + 18=0\))
Then \(a = 4\), \(b=-2\), \(c = 9\)
Quadratic formula: \(n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{2\pm\sqrt{4 - 144}}{8}=\frac{2\pm\sqrt{-140}}{8}=\frac{2\pm2i\sqrt{35}}{8}=\frac{1\pm i\sqrt{35}}{4}\), not 37.

Wait, maybe the original equation is \(8n^{2}-4n=18\) is \(8n^{2}-4n=-18\), so \(8n^{2}-4n + 18=0\), and there is a miscalculation in \(c\). If \(c=-9\) instead of 9, then \(8n^{2}-4n-9 = 0\), discriminant is \(16+288 = 304\), no.

Wait, let's start over with the equation \(8n^{2}-4n=18\)
Subtract 18: \(8n^{2}-4n - 18=0\)
Divide by 2: \(4n^{2}-2n - 9=0\)
\(a = 4\), \(b=-2\), \(c=-9\)
Discriminant: \(b^{2}-4ac=(-2)^{2}-4\times4\times(-9)=4 + 144 = 148\)
\(\sqrt{148}=\sqrt{4\times37}=2\sqrt{37}\)
So \(n=\frac{2\pm2\sqrt{37}}{8}=\frac{1\pm\sqrt{37}}{4}\)
These are real solutions, and the options have \(\frac{1\pm\sqrt{37}}{4}\) as options (the fourth and fifth options). But the options with \(i\) are complex. So maybe there was a sign error in the original equation. If the equation was \(8n^{2}-4n=-18\), then \(8n^{2}-4n + 18=0\), \(a = 8\), \(b=-4\), \(c = 18\)
Discriminant: \(16-576=-560\), \(\sqrt{-560}=4i\sqrt{35}\), not 37.

Alternatively, maybe the equation is \(4n^{2}-2n + 9=0\), and discriminant is \(4-144=-140\), \(\sqrt{-140}=2i\sqrt{35}\), no.

Wait, the options with \(i\) are \(\frac{1\pm i\sqrt{37}}{4}\). Let's see, if discriminant is \(-144 + 4=-140\) no, but if \(a = 1\), \(b=-1\), \(c=-9\), discriminant is \(1 + 36 = 37\), but that's not our case.

Wait, maybe the original equation is \(n^{2}-2n + 9=0\), then discriminant is \(4-36=-32\), no.

Alternatively, maybe the equation is \(4n^{2}-2n + 9=0\), and we made a mistake in \(a\), \(b\), \(c\). Let's take the option \(\frac{1\pm i\sqrt{37}}{4}\), let's square the numerator: \((1\pm i\sqrt{37})^{2}=1\pm2i\sqrt{37}-37=-36\pm2i\sqrt{37}\), and denominator is 16. Not helpful.

Wait, let's use the quadratic formula for the equation \(4n^{2}-2n + 9=0\) (assuming \(8n^{2}-4n + 18=0\)):

\(n=\frac{2\pm\sqrt{4-144}}{8}=\frac{2\pm\sqrt{-140}}{8}=\frac{2\pm2i\sqrt{35}}{8}=\frac{1\pm i\sqrt{35}}{4}\)

But the options have \(\sqrt{37}\), so maybe there is a typo in the problem, and the equation is \(8n^{2}-4n=-18\) with \(c = - 9\) instead of 9, so \(8n^{2}-4n-9=0\), no, discriminant is \(16 + 288=304\).

Alternatively, maybe the original equation is \(8n^{2}-4n=18\) and the options with \(i\) are wrong, and the correct real solutions are \(\frac{1\pm\sqrt{37}}{4}\). But the options have both real and complex solutions.

Wait, let's recalculate the discriminant for \(8n^{2}-4n-18=0\):

\(a = 8\), \(b=-4\), \(c=-18\)

\(b^{2}-4ac=(-4)^{2}-4\times8\times(-18)=16 + 576=592\)

\(\sqrt{592}=\sqrt{16\times37}=4\sqrt{37}\)

Then \(n=\frac{4\pm4\sqrt{37}}{16}=\frac{1\pm\sqrt{37}}{4}\)

Ah! Here we go. I made a mistake earlier when dividing by 2. If we don't divide by 2, \(a = 8\), \(b=-4\), \(c=-18\)

So \(n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{4\pm\sqrt{(-4)^{2}-4\times8\times(-18)}}{2\times8}=\frac{4\pm\sqrt{16 + 576}}{16}=\frac{4\pm\sqrt{592}}{16}=\frac{4\pm4\sqrt{37}}{16}=\frac{1\pm\sqrt{37}}{4}\)

So the real solutions are \(\frac{1\pm\sqrt{37}}{4}\), which are the fourth and fifth options. But the options with \(i\) are complex, which would be the case if the discriminant is negative. So maybe there was a sign error in the original equation, and it's \(8n^{2}-4n=-18\), leading to complex solutions. But based on the given equation \(8n^{2}-4n = 18\), the real solutions are \(\frac{1\pm\sqrt{37}}{4}\)

Answer:

\(\frac{1-\sqrt{37}}{4}\), \(\frac{1+\sqrt{37}}{4}\) (corresponding to the fourth and fifth options: \(\frac{1-\sqrt{37}}{4}\), \(\frac{1+\sqrt{37}}{4}\))