Math  /  Algebra

Questionz(z+8)=18z(z+8)=-18
The solution set is \square }\}

Studdy Solution

STEP 1

1. The equation z(z+8)=18 z(z+8) = -18 is a quadratic equation.
2. Solving the equation involves expanding and rearranging terms to form a standard quadratic equation.
3. The quadratic formula or factoring will be used to find the solutions.

STEP 2

1. Expand and rearrange the equation.
2. Solve the quadratic equation.
3. Verify the solutions.

STEP 3

Expand the left-hand side of the equation:
z(z+8)=z2+8z z(z + 8) = z^2 + 8z

STEP 4

Rearrange the equation to form a standard quadratic equation by moving all terms to one side:
z2+8z+18=0 z^2 + 8z + 18 = 0

STEP 5

Attempt to factor the quadratic equation. We look for two numbers that multiply to 18 18 and add to 8 8 . Since factoring is not straightforward, we use the quadratic formula:
z=b±b24ac2a z = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
where a=1 a = 1 , b=8 b = 8 , and c=18 c = 18 .

STEP 6

Calculate the discriminant b24ac b^2 - 4ac :
b24ac=824118=6472=8 b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 18 = 64 - 72 = -8

STEP 7

Since the discriminant is negative, the solutions are complex. Calculate the solutions using the quadratic formula:
z=8±82 z = \frac{{-8 \pm \sqrt{{-8}}}}{2}
z=8±2i22 z = \frac{{-8 \pm 2i\sqrt{2}}}{2}
z=4±i2 z = -4 \pm i\sqrt{2}

STEP 8

Verify the solutions by substituting back into the original equation. Since the solutions are complex, they are valid as the equation involves real coefficients.
The solution set is:
{4+i2,4i2} \{-4 + i\sqrt{2}, -4 - i\sqrt{2}\}
The solution set is:
{4+i2,4i2} \boxed{\{-4 + i\sqrt{2}, -4 - i\sqrt{2}\}}

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