Find zeros of quadratic equation

AI Thread Summary
To find the zeros of the quadratic equation P(x) = 4(x+7)² + (x+7) - 3, a substitution is made where u = x + 7, transforming the equation into f(u) = 4u² + u - 3. After factoring this quadratic equation, the next step involves solving for u, which leads to specific values. Once the values for u are determined, back substitution is necessary to find x using the equation x = u - 7. This method effectively allows for the calculation of the zeros of the original quadratic equation.
Torshi
Messages
118
Reaction score
0

Homework Statement


Find the zeros of the quadratic equation?


Homework Equations



P(x)=4(x+7)^2+(x+7)-3

The Attempt at a Solution



I substituted and made it 4u^2+u-3 ?? Then after that I factor then back substitute?? I'm not sure how to back substitute tho.
 
Physics news on Phys.org
Think about what u is equal to when you make the substitution. If you have that and the value for u, finding x should be straightforward.
 
Torshi said:
I'm not sure how to back substitute tho.

What's the equation for u in terms of x? Once you have this, solve for x in the equation and plug in the values you got for u.
 
You did a change of variables. You substituted u for x where u = x+7 to get the equation

f(u) \, = \, 4u^{2} \, + \, u \, - \, 3 \, = \, 0

solve for u in the above expression then use u = x + 7 to find x for each u.

i.e. since u = x+7 then x = u - 7 gives you x.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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
Back
Top