Thread: 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.

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

Note:
Be sure you assure that your solution

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.

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
Solve using quadratic formula.

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:


@Wilmer:

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.

Hello, benny92000!

Whenever possible, I would avoid fractions.

Subtract 3 from both sides: .

Square both sides: .

. . . . . . . . . . . . . . . . . .

Simplify:. a . . . .

Factor:. - . . . . . .


And we have: .

. . and both answers check out!

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

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