1. Solving for X

Here is the problem I have: -7 x [sqr root of (3x-8)] + 3 = -4x + 7. The answer I am looking for is 8. I first subtracted 3 then divided the right side by -7. Then I squared both sides and attempted to simplify. I didn't come up with the right answer.

2. Re: Solving for X

$-7\sqrt{3x-8} = -4x+4$

$\sqrt{3x-8} = -\dfrac{(-4x+4)}{7} = \dfrac{4}{7}(x-1)$

$3x-8 = \left(\dfrac{4}{7}\right)^2 \cdot (x-1)^2$

Do you follow up to there? If so, what do you get on the RHS?

3. Re: Solving for X

Note:
Be sure you assure that your solution $x\geq\frac{8}{3}$

4. Re: Solving for X

Yes, I follow up to there. I personally didn't break up the denominator into a fraction as you did. I get 16/49 times x^2 - 2x + 1 on the RHS. I tried distributing the fraction over and I got the wrong answer again... I don't know what I'm doing wrongly. I've worked this at least 10 times, coming with the quadratic and plugging 8 in, and haven't been successful.

5. Re: Solving for X

Well, if you're having problems with fractions, then just square both sides:
[-7SQRT(3x - 8)] squared = 49(3x - 8)
[-4x + 4] squared = 16x^2 + 16x + 16
So:
147x - 392 = 16x^2 + 16x + 16

6. Re: Solving for X

Originally Posted by benny92000
I personally didn't break up the denominator into a fraction as you did. I get 16/49 times x^2 - 2x + 1 on the RHS.
This is correct for the RHS, multiply both sides with 49 so you're 'losing' the denominator in the RHS:
$49(3x-8)=16(x^2-2x+1)$

@Wilmer:
$(-4x+4)^2=16x^2-32x+16$

7. Re: Solving for X

I came up with the right answer. I had a binomial and I was squaring each term individually rather than the binomial as a whole... -.- Thanks guys.

8. Re: Solving for X

Hello, benny92000!

$-7\sqrt{3x-8}+ 3 \:=\: -4x + 7$

Whenever possible, I would avoid fractions.

Subtract 3 from both sides: . $-7\sqrt{3x-8} \:=\:-4x + 4$

Square both sides: . $\bigg[-7\sqrt{3x-8}\bigg]^2 \:=\:\bigg[-4x + 4\bigg]^2$

. . . . . . . . . . . . . . . . . . $49(3x-8) \:=\:16x^2 - 32x + 16$

Simplify:. a . . . . $16x^2 - 179x + 408 \:=\:0$

Factor:. - . . . . . . $(x-8)(16x-51) \:=\:0$

And we have: . $\begin{Bmatrix}x - 8 \:=\: 0 & \Rightarrow & x \:=\: 8 \\ 16x - 51 \:=\: 0 & \Rightarrow & x \:=\: \frac{51}{16} \end{Bmatrix}$

. . and both answers check out!

9. Re: Solving for X

sorry thought i saw something but now i can't delete this post.could anyone do it?

10. Re: Solving for X

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