What is the Solution for x When \( (4+3x)(6+3x) = 24 \)?

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The equation \( (4+3x)(6+3x) = 24 \) represents a quadratic function, which can be graphed to find intersections with the line \( y = 24 \). One user mentions that the vertex of the parabola is approximately at \( x = -1.7 \) and \( y = -1 \). Another suggests using a graphing application on a computer, hinting at Maple 8 for graphing purposes. A simpler method is proposed by expanding the equation to \( 9x^2 + 30x = 0 \), which can be factored and solved without a calculator. The discussion emphasizes both graphical and algebraic approaches to solving the equation.
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Graphing calc. died :( i need help

could some 1 but this in a calc and tell me the intersection when y is 24?

(4+3x)(6+3x)=24

thanks :shy:
 
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according to my calculator (just started using it a few days ago), it's a parabola (sp?) and the vertex is about x = -1.7, y = -1. that's just with me using the trace function (i don't know how to get the vertex the normal way). hope this helps.

~Amy
 
physicsgal said:
according to my calculator (just started using it a few days ago), it's a parabola (sp?) and the vertex is about x = -1.7, y = -1. that's just with me using the trace function (i don't know how to get the vertex the normal way). hope this helps.

~Amy

did you put (4+3x)(6+3x) into y1, and 24 into y2, after that make window bigger than 2end trace 5, then you get it.
 
Well, you obviously are sitting at a computer if you can post on PF that your graphing calculator died. Quiz question -- what application on your PC can be used for graphing? Hint -- it rhymes with Extra Large.
 
hm..sounds like Extra Large...Maple 8!
 
feod2003 said:
could some 1 but this in a calc and tell me the intersection when y is 24?

(4+3x)(6+3x)=24

thanks :shy:

Why would you need a calculator of any kind to do that? Multiply it out: 9x2+ 30x+ 24= 24 so 9x2+ 30x= 0. That's easy to factor and solve.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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