xยฒ3x3x9x = 5b/2 = 3x = 5b/2 = 3
Splitting b/2 term3
Completed Corner Area (b/2)ยฒ9
Calculated Roots (x) x โ‰ˆ 2.0, x โ‰ˆ -8.0

Variable Adjuster

Visual Variable Value (x)5
38
Coefficient b (bx) - Even values preferred6
210
Constant c (Target Area)16
140

Completing the Square

SQCOM

Completing the square is a geometric technique to solve quadratic equations. We model x^2 as a square of side x, and bx as two side rectangles of width b/2. Adding a small corner square of area (b/2)^2 completes the shape into a larger perfect square.

x2+bx+(b2)2=(x+b2)2x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2

Whiteboard Solver Steps

Step 1

Original Equation

We begin with our quadratic equation in the form x2+bx=cx^2 + bx = c, where we have an unknown square of side length xx (area x2x^2) plus bb rectangular strips of width 11 and height xx (total area 6x),equatingtoatargetareaof16.), equating to a target area of 16.

x2+6x=16x^2 + 6x = 16
Step 2

Split the Linear Coefficient (Geometric Alignment)

To maintain symmetry and build towards a perfect square shape, we divide the linear coefficient (b=6b = 6) in half, giving b/2=3b/2 = 3. Geometrically, we split the area bxbx into two equal rectangles of dimensions xร—3x \times 3 and place one on the right side of the main x2x^2 square, and the other at the bottom.

b2=62=3\frac{b}{2} = \frac{6}{2} = 3
Step 3

Calculate the Completing Area (Corner Square)

Observe the empty corner that remains at the bottom-right. The dimensions of this missing space are exactly b/2ร—b/2b/2 \times b/2 (or 3 \times 3).Tofillthiscornerandformasingle,largeperfectsquareofside). To fill this corner and form a single, large perfect square of side (x + b/2),wecalculateitsarea:, we calculate its area: (3)^2 = 9..

(b2)2=32=9\left(\frac{b}{2}\right)^2 = 3^2 = 9
Step 4

Add Completing Term to Both Sides (Algebretic Balance)

To preserve algebraic balance, we add the corner area (99) to both sides of the equation. The left side is now a perfect trinomial square: x2+6x+9=(x+3)2x^2 + 6x + 9 = (x + 3)^2. The right side becomes c+(b/2)2=16+9=25c + (b/2)^2 = 16 + 9 = 25.

x2+6x+9=16+9(x+3)2=25\begin{aligned}x^2 + 6x + 9 = 16 + 9 \\ (x + 3)^2 = 25\end{aligned}
Step 5

Take the Square Root

To isolate xx, we take the square root of both sides. This yields two possibilities: x+3=25x + 3 = \sqrt{25} and x+3=โˆ’25x + 3 = -\sqrt{25}.

x+3=ยฑ25โ‰ˆยฑ5.00x + 3 = \pm \sqrt{25} \approx \pm 5.00
Step 6

Solve for x (Find Roots)

Subtract b/2=3b/2 = 3 from both sides to find the solutions. The two real roots represent the values of xx where the equation balances, giving x1โ‰ˆ2x_1 \approx 2 and x2โ‰ˆโˆ’8x_2 \approx -8.

x1=โˆ’3+5.00=2x2=โˆ’3โˆ’5.00=โˆ’8\begin{aligned}x_1 = -3 + 5.00 = 2 \\ x_2 = -3 - 5.00 = -8\end{aligned}