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人気のある代数 >

solvefor t,St+1=t^2+S

解

解く

t, St + 1 = t^{2} + S

解

t = \frac{S + S ^{2} - 4 ( S - 1 )}{2}, t = \frac{S - S ^{2} - 4 ( S - 1 )}{2}

解答ステップ

St + 1 = t^{2} + S

辺を交換する

t^{2} + S = St + 1

1

を左側に移動します

t^{2} + S - 1 = St

St

を左側に移動します

t^{2} + S - 1 - St = 0

標準的な形式で書く

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

t^{2} - St + S - 1 = 0

解くとthe二次式

t_{1, 2} = \frac{- ( - S ) \pm ( - S ) ^{2} - 4 \cdot 1 \cdot ( S - 1 )}{2 \cdot 1}

簡素化

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

t_{1, 2} = \frac{- ( - S ) \pm S ^{2} - 4 ( S - 1 )}{2 \cdot 1}

解を分離する

t_{1} = \frac{- ( - S ) + S ^{2} - 4 ( S - 1 )}{2 \cdot 1}, t_{2} = \frac{- ( - S ) - S ^{2} - 4 ( S - 1 )}{2 \cdot 1}

t = \frac{- ( - S ) + S ^{2} - 4 ( S - 1 )}{2 \cdot 1} : \frac{S + S ^{2} - 4 ( S - 1 )}{2}

t = \frac{- ( - S ) - S ^{2} - 4 ( S - 1 )}{2 \cdot 1} : \frac{S - S ^{2} - 4 ( S - 1 )}{2}

二次equationの解:

t = \frac{S + S ^{2} - 4 ( S - 1 )}{2}, t = \frac{S - S ^{2} - 4 ( S - 1 )}{2}

グラフ

人気の例

3 t (t + 2) = 0

solvefor x,r^2=(y-r)^2+(x/2)^2

so l v e f or x, r^{2} = (y - r)^{2} + (\frac{x}{2})^{2}

88 = 151 x - 16 x^{2}

- x^{2} - 2 x = - 47

x^{2} + 10 x + 169 = 0