How do you solve $\sqrt{10 - 3 b} = b$ $\sqrt { 10- 3b } = b$ ?

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Answer 1 · 2017-12-19T17:49:40

b=2

$\sqrt{10 - 3 b} = b$ $sqrt(10-3b)=b$

Just square bost side os the equality:

$(10 - 3 b) = b^{2}$ $(10-3b)=b^2$

$0 = b^{2} + 3 b - 10$ $0=b^2+3b-10$

$b = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot 1 \cdot (- 10)}}{2}$ $b=(-3+-sqrt(3^2-4*1*(-10)))/2$

$b = \frac{- 3 \pm \sqrt{49}}{2}$ $b=(-3+-sqrt(49))/2$

$b = \frac{- 3 \pm 7}{2}$ $b=(-3+-7)/2$

$b = \frac{- 3 + 7}{2}$ $b=(-3+7)/2$ ou $b = \frac{- 3 - 7}{2}$ $b=(-3-7)/2$

$b = 2$ $b=2$ ou $b = - 5$ $b=-5$

You must determine the dominion of this expression:

since there is a square root and the solution must be a posiive number

$10 - 3 b > 0$ $10-3b>0$ and $b > 0$ $b>0$

$10 > 3 b$ $10>3b$

$b < \frac{10}{3}$ $b<10/3$ and $b > 0$ $b>0$

Concernig this, only b=2 is valid

How do you solve √10−3b=b\sqrt { 10- 3b } = b?