Question

now fill in the correct values.
be extra careful with the signs. take note of what
signs are already given to you.
$x^2 -8x + 18 = 0$
$x = \frac{- (\square) \pm \sqrt{(\square)^2 - \square (\square)(\square)}}{\square (\square)}$

Explanation:

Step1: Recall Quadratic Formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).

Step2: Identify \(a\), \(b\), \(c\)

For the equation \(x^2 - 8x + 18 = 0\), we have \(a = 1\), \(b=-8\), \(c = 18\).

Step3: Substitute into Formula

Substitute \(a = 1\), \(b=-8\), \(c = 18\) into the quadratic formula:

• The numerator's first part: \(-b=-(-8)=8\) (but in the given format, the first box is for \(b\) as per the formula's \(-b\), wait no—the formula is \(\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), so the first box (inside the \(-(\square)\)) is \(b\), which is \(-8\)? Wait, no: the formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), so the first term in the numerator is \(-b\), so if \(b=-8\), then \(-b = 8\)? Wait, no, the equation is \(x^2-8x + 18=0\), so \(a = 1\), \(b=-8\)? Wait, no: standard form is \(ax^2+bx + c = 0\), so \(x^2-8x + 18=0\) means \(a = 1\), \(b=-8\)? Wait, no, \(x^2+bx + c\), so \(b=-8\), \(a = 1\), \(c = 18\). Then the quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(18)}}{2(1)}\). Wait, but the given format is \(x=\frac{-(\square)\pm\sqrt{(\square)^2 - \square(\square)(\square)}}{\square(\square)}\). Let's map:
• The first box (inside \(-(\square)\)): this is \(b\), because the formula is \(-b\), so \(b=-8\), so \(-b=-(-8)=8\), but in the format, the first box is the value inside \(-(\square)\), so that's \(b\) which is \(-8\)? Wait, no, let's look at the format: \(x=\frac{-(\square)\pm\sqrt{(\square)^2 - \square(\square)(\square)}}{\square(\square)}\). So:
• First box (inside \(-(\square)\)): \(b\), which is \(-8\)? Wait, no, the formula is \(-b\), so if \(b=-8\), then \(-b = 8\), but the format has \(-(\square)\), so \(\square\) is \(b\), so \(\square=-8\) (because \(-(-8)=8\)).
• The next box (inside the square root, the \((\square)^2\)): this is \(b\), so \(b=-8\), so \(\square=-8\).
• Then the next part: \(4ac\) is \(4\times a\times c\), so \(a = 1\), \(c = 18\), so the boxes for \(4ac\) are \(4\), \(a = 1\), \(c = 18\) (so the three boxes: \(\square(\square)(\square)\) is \(4\), \(1\), \(18\)).
• The denominator: \(2a\), so \(2\) and \(a = 1\), so the two boxes: \(\square(\square)\) is \(2\), \(1\).

Wait, let's re - express:

Given the equation \(x^2-8x + 18=0\), \(a = 1\), \(b=-8\), \(c = 18\).

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

So:

• The first box (inside \(-(\square)\)): \(b=-8\) (because \(-b=-(-8)=8\), but the format is \(-(\square)\), so \(\square = b=-8\))
• The second box (inside the square root, \((\square)^2\)): \(b=-8\)
• The next three boxes (for \(4ac\)): \(4\), \(a = 1\), \(c = 18\) (so \(4\times1\times18\))
• The denominator boxes: \(2\), \(a = 1\) (so \(2\times1\))

So filling in:

\(x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(18)}}{2(1)}\)

So the first box (inside \(-(\square)\)): \(-8\) (wait, no: \(-b\) when \(b = - 8\) is \(8\), but the format is \(-(\square)\), so \(\square=b=-8\), because \(-(-8)=8\))

The second box (inside \(\sqrt{(\square)^2}\)): \(-8\)

The next three boxes (for \(4ac\)): \(4\), \(1\), \(18\)

The denominator boxes: \(2\), \(1\)

Answer:

First box: \(-8\), Second box: \(-8\), Third box: \(4\), Fourth box: \(1\), Fifth box: \(18\), Sixth box: \(2\), Seventh box: \(1\)

(In the given format:

\(x=\frac{-(\boldsymbol{-8})\pm\sqrt{(\boldsymbol{-8})^2-\boldsymbol{4}(\boldsymbol{1})(\boldsymbol{18})}}{\boldsymbol{2}(\boldsymbol{1})}\)