Answer :
Certainly! Let's solve the equation [tex]\( d^2 - 12d + 36 = 4 \)[/tex] step-by-step using the square roots property.
1. First, bring the equation to standard form:
[tex]\[ d^2 - 12d + 36 - 4 = 0 \][/tex]
Which simplifies to:
[tex]\[ d^2 - 12d + 32 = 0 \][/tex]
2. Next, notice that the equation is a perfect square trinomial:
[tex]\[ (d - 6)^2 = 4 \][/tex]
3. Apply the square root property:
[tex]\[ d - 6 = \pm 2 \][/tex]
This gives us two equations to solve:
[tex]\[ \begin{cases} d - 6 = 2 \\ d - 6 = -2 \end{cases} \][/tex]
4. Solve each equation individually:
[tex]\[ \begin{cases} d - 6 = 2 & \Rightarrow d = 2 + 6 = 8 \\ d - 6 = -2 & \Rightarrow d = -2 + 6 = 4 \end{cases} \][/tex]
So, the solutions to the equation [tex]\( d^2 - 12d + 36 = 4 \)[/tex] are:
[tex]\[ d = 8, 4 \][/tex]
1. First, bring the equation to standard form:
[tex]\[ d^2 - 12d + 36 - 4 = 0 \][/tex]
Which simplifies to:
[tex]\[ d^2 - 12d + 32 = 0 \][/tex]
2. Next, notice that the equation is a perfect square trinomial:
[tex]\[ (d - 6)^2 = 4 \][/tex]
3. Apply the square root property:
[tex]\[ d - 6 = \pm 2 \][/tex]
This gives us two equations to solve:
[tex]\[ \begin{cases} d - 6 = 2 \\ d - 6 = -2 \end{cases} \][/tex]
4. Solve each equation individually:
[tex]\[ \begin{cases} d - 6 = 2 & \Rightarrow d = 2 + 6 = 8 \\ d - 6 = -2 & \Rightarrow d = -2 + 6 = 4 \end{cases} \][/tex]
So, the solutions to the equation [tex]\( d^2 - 12d + 36 = 4 \)[/tex] are:
[tex]\[ d = 8, 4 \][/tex]