Answer :
To solve the quadratic equation \( v^2 + 8v - 3 = 0 \) by completing the square, follow these steps:
1. Rearrange the Equation:
Start with the quadratic equation:
[tex]\[ v^2 + 8v - 3 = 0 \][/tex]
Move the constant term to the right side of the equation:
[tex]\[ v^2 + 8v = 3 \][/tex]
2. Complete the Square:
To complete the square, we need to form a perfect square trinomial on the left side. Take the coefficient of \( v \), divide it by 2, and square the result:
[tex]\[ \left(\frac{8}{2}\right)^2 = 4^2 = 16 \][/tex]
Add and subtract this value inside the equation to maintain equality:
[tex]\[ v^2 + 8v + 16 - 16 = 3 \][/tex]
Simplify to group the perfect square trinomial on the left:
[tex]\[ (v + 4)^2 - 16 = 3 \][/tex]
Now, isolate the perfect square trinomial:
[tex]\[ (v + 4)^2 = 3 + 16 \][/tex]
[tex]\[ (v + 4)^2 = 19 \][/tex]
3. Solve for \( v \):
Take the square root of both sides:
[tex]\[ v + 4 = \pm \sqrt{19} \][/tex]
Solve for \( v \) by isolating it on one side of the equation:
[tex]\[ v = -4 \pm \sqrt{19} \][/tex]
Thus, the solutions to the equation \( v^2 + 8v - 3 = 0 \) are in the form:
[tex]\[ v = a \pm \sqrt{b} \][/tex]
where \( a = -4 \) and \( b = 19 \).
Therefore, the solutions are:
[tex]\[ v = -4 \pm \sqrt{19} \][/tex]
1. Rearrange the Equation:
Start with the quadratic equation:
[tex]\[ v^2 + 8v - 3 = 0 \][/tex]
Move the constant term to the right side of the equation:
[tex]\[ v^2 + 8v = 3 \][/tex]
2. Complete the Square:
To complete the square, we need to form a perfect square trinomial on the left side. Take the coefficient of \( v \), divide it by 2, and square the result:
[tex]\[ \left(\frac{8}{2}\right)^2 = 4^2 = 16 \][/tex]
Add and subtract this value inside the equation to maintain equality:
[tex]\[ v^2 + 8v + 16 - 16 = 3 \][/tex]
Simplify to group the perfect square trinomial on the left:
[tex]\[ (v + 4)^2 - 16 = 3 \][/tex]
Now, isolate the perfect square trinomial:
[tex]\[ (v + 4)^2 = 3 + 16 \][/tex]
[tex]\[ (v + 4)^2 = 19 \][/tex]
3. Solve for \( v \):
Take the square root of both sides:
[tex]\[ v + 4 = \pm \sqrt{19} \][/tex]
Solve for \( v \) by isolating it on one side of the equation:
[tex]\[ v = -4 \pm \sqrt{19} \][/tex]
Thus, the solutions to the equation \( v^2 + 8v - 3 = 0 \) are in the form:
[tex]\[ v = a \pm \sqrt{b} \][/tex]
where \( a = -4 \) and \( b = 19 \).
Therefore, the solutions are:
[tex]\[ v = -4 \pm \sqrt{19} \][/tex]
