What are the solutions to the equation log_{10}(10)(a(a+3))=1 ?

The solutions to the equation log_{10}(10)(a(a+3))=1 are a=(-3+sqrt(13))/2 ,a=(-3-sqrt(13))/2

Find the zeros of log_{10}(10)(a(a+3))=1

The zeros of log_{10}(10)(a(a+3))=1 are a=(-3+sqrt(13))/2 ,a=(-3-sqrt(13))/2

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lo g_{10} (10) (a (a + 3)) = 1

a = \frac{- 3 + 13}{2}, a = \frac{- 3 - 13}{2}

Decimal Notation

a = 0.30277 \dots, a = - 3.30277 \dots

Solution steps

lo g_{10} (10) (a (a + 3)) = 1

Expand

lo g_{10} (10) (a (a + 3)) : a^{2} + 3 a

a^{2} + 3 a = 1

Move

1

to the left side

a^{2} + 3 a - 1 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

a_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 3 + 13}{2 \cdot 1}, a_{2} = \frac{- 3 - 13}{2 \cdot 1}

a = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

a = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

a = \frac{- 3 + 13}{2}, a = \frac{- 3 - 13}{2}

x^{2} - (k + 2) x + (k + 5) = 0

x, x^{2} (y - z) p + y^{2} (z - x) q = z^{2} (x - y)

x^{2} - 2 (a^{2} + b^{2}) x + (a^{2} - b^{2})^{2} = 0

a, a^{2} + b^{2} + 2 (3 a + 2 b) + 13 = 0

y, x = y^{2} + 7 z^{2}

What are the solutions to the equation log_{10}(10)(a(a+3))=1 ?
The solutions to the equation log_{10}(10)(a(a+3))=1 are a=(-3+sqrt(13))/2 ,a=(-3-sqrt(13))/2
Find the zeros of log_{10}(10)(a(a+3))=1
The zeros of log_{10}(10)(a(a+3))=1 are a=(-3+sqrt(13))/2 ,a=(-3-sqrt(13))/2