How would yous solve $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$ ?

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How would yous solve $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$ ?

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Answer 1 · 2018-05-15T15:01:49

$x = 0.129$ $x=0.129$

Explanation:

In the equation $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$ we can only have $3 x + 1 \geq 0$ $3x+1>=0$ and $8 x - 1 \geq 0$ $8x-1>=0$ i.e. $x \geq - \frac{1}{3}$ $x>=-1/3$ and $x \geq \frac{1}{8}$ $x>=1/8$ i.e. $x \geq \frac{1}{8}$ $x>=1/8$ or $x \geq 0.125$ $x>=0.125$ .

To solve the equation $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$

squaring each side, we get

$3 x + 1 - 2 \sqrt{3 x + 1} + 1 = 8 x - 1$ $3x+1-2sqrt(3x+1)+1=8x-1$

now take irrational portion on the left and we get

$- 2 \sqrt{3 x + 1} = 8 x - 1 - 3 x - 1 - 1 = 5 x - 3$ $-2sqrt(3x+1)=8x-1-3x-1-1=5x-3$

squaring again $4 (3 x + 1) = 25 x^{2} - 30 x + 9$ $4(3x+1)=25x^2-30x+9$

or $25 x^{2} - 30 x + 9 = 12 x + 4$ $25x^2-30x+9=12x+4$

or $25 x^{2} - 42 x + 5 = 0$ $25x^2-42x+5=0$

or $x = \frac{42 \pm \sqrt{42^{2} - 500}}{50} = \frac{42 \pm \sqrt{1264}}{50} = \frac{42 \pm 35.553}{50}$ $x=(42+-sqrt(42^2-500))/50=(42+-sqrt1264)/50=(42+-35.553)/50$

i.e. $x = 0.129$ $x=0.129$ or $1.551$ $1.551$

However on checking $1.551$ $1.551$ is not found to be a solution and hence only answer is $x = 0.129$ $x=0.129$

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How would yous solve $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$ ?

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Explanation:

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How would yous solve sqrt(3x+1)-1=sqrt(8x-1)√3x+1−1=√8x−1sqrt(3x+1)-1=sqrt(8x-1)?

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How would yous solve $\sqrt{3 x + 1} - 1 = \sqrt{8 x - 1}$ $sqrt(3x+1)-1=sqrt(8x-1)$ ?