Math  /  Algebra

QuestionFind the zeros of h(x)=5x2+40h(x)=5 x^{2}+40.
The zeros are x=x= \square and x=x= \square \square.

Studdy Solution

STEP 1

1. The function h(x)=5x2+40 h(x) = 5x^2 + 40 is a quadratic function.
2. The zeros of a function are the values of x x for which h(x)=0 h(x) = 0 .
3. Solving for the zeros involves setting the function equal to zero and solving the resulting equation.

STEP 2

1. Set the function equal to zero.
2. Solve the resulting equation for x x .
3. Simplify the solutions to find the zeros.

STEP 3

Set the function equal to zero. This means we want to solve the equation:
5x2+40=0 5x^2 + 40 = 0

STEP 4

Isolate the x2 x^2 term. Start by subtracting 40 from both sides of the equation:
5x2+4040=040 5x^2 + 40 - 40 = 0 - 40 5x2=40 5x^2 = -40

STEP 5

Divide both sides by 5 to solve for x2 x^2 :
5x25=405 \frac{5x^2}{5} = \frac{-40}{5} x2=8 x^2 = -8

STEP 6

Take the square root of both sides to solve for x x . Remember that taking the square root of a negative number involves imaginary numbers:
x=±8 x = \pm \sqrt{-8}

STEP 7

Simplify the square root of the negative number. Recall that 1=i \sqrt{-1} = i , where i i is the imaginary unit:
x=±8 x = \pm \sqrt{-8} x=±81 x = \pm \sqrt{8} \cdot \sqrt{-1} x=±8i x = \pm \sqrt{8} \cdot i

STEP 8

Simplify 8 \sqrt{8} . Since 8=4×2 8 = 4 \times 2 and 4=2 \sqrt{4} = 2 , we have:
x=±22i x = \pm 2\sqrt{2} \cdot i
The zeros of the function h(x)=5x2+40 h(x) = 5x^2 + 40 are:
x=22iandx=22i x = 2\sqrt{2}i \quad \text{and} \quad x = -2\sqrt{2}i

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