What are the solutions to the equation k(x)=4x^2+1,k(-3) ?

The solutions to the equation k(x)=4x^2+1,k(-3) are x=(k+sqrt(k^2-16))/8 ,x=(k-sqrt(k^2-16))/8

Find the zeros of k(x)=4x^2+1,k(-3)

The zeros of k(x)=4x^2+1,k(-3) are x=(k+sqrt(k^2-16))/8 ,x=(k-sqrt(k^2-16))/8

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k (x) = 4 x^{2} + 1, k (- 3)

x = \frac{k + k ^{2} - 16}{8}, x = \frac{k - k ^{2} - 16}{8}

Solution steps

k (x) = 4 x^{2} + 1

Refine

k x = 4 x^{2} + 1

Switch sides

4 x^{2} + 1 = k x

Move

k x

to the left side

4 x^{2} + 1 - k x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - k x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - k ) \pm ( - k ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

Simplify

(- k)^{2} - 4 \cdot 4 \cdot 1 : k^{2} - 16

x_{1, 2} = \frac{- ( - k ) \pm k ^{2} - 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - k ) + k ^{2} - 16}{2 \cdot 4}, x_{2} = \frac{- ( - k ) - k ^{2} - 16}{2 \cdot 4}

x = \frac{- ( - k ) + k ^{2} - 16}{2 \cdot 4} : \frac{k + k ^{2} - 16}{8}

x = \frac{- ( - k ) - k ^{2} - 16}{2 \cdot 4} : \frac{k - k ^{2} - 16}{8}

The solutions to the quadratic equation are:

x = \frac{k + k ^{2} - 16}{8}, x = \frac{k - k ^{2} - 16}{8}

- x^{2} = 8 x + 1

(x - 8) (x + 8) = 0

- 4.9 t^{2} + 30 t = 0

z^{2} - 4 z + 13 = 0

8 x^{2} - 2 x = 7

What are the solutions to the equation k(x)=4x^2+1,k(-3) ?
The solutions to the equation k(x)=4x^2+1,k(-3) are x=(k+sqrt(k^2-16))/8 ,x=(k-sqrt(k^2-16))/8
Find the zeros of k(x)=4x^2+1,k(-3)
The zeros of k(x)=4x^2+1,k(-3) are x=(k+sqrt(k^2-16))/8 ,x=(k-sqrt(k^2-16))/8