anonymous
  • anonymous
Find the possible values of b for which the equation 9x^2 + bx + 7 = o has roots which differ by 2.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This time round i hope someone can post the sum of roots and product of roots method *please*
BAdhi
  • BAdhi
if the quadratic equation is, $$ax^2+bx+c=0$$and roots are $$\alpha, \; \beta$$ $$\alpha=\frac{-b-\sqrt{\Delta}}{2a},\quad \beta=\frac{-b+\sqrt{\Delta}}{2a}\\ |\alpha-\beta|=\frac{\sqrt{\Delta}}{2a} \quad\text{(for $a>0$)}$$where $$\Delta=b^2-4ac$$
anonymous
  • anonymous
I have not learnt modulus yet

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anonymous
  • anonymous
going to soon
BAdhi
  • BAdhi
if you want only from sum and products of the roots, use, $$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$$
anonymous
  • anonymous
Can you show me that steps please?
anonymous
  • anonymous
to derive out the possible values of b
anonymous
  • anonymous
Yours sincerely, very much appreciate it.
BAdhi
  • BAdhi
you know that $$(\alpha-\beta)^2=2^2=4\\ \alpha+\beta=-\frac ba=-\frac b 9\\ \alpha\beta=\frac ca=\frac 7 9$$ Substituting values to previous given identity, $$4=\left(\frac{-b}{9}\right)^2-4\left(\frac 79 \right)$$ now solve for b. Hope this is helpful
anonymous
  • anonymous
how did you get (alpha - beta) square ?
BAdhi
  • BAdhi
In the problem they have given that the difference between the roots is 2. I have taken the roots as alpha and beta Thus we can imply, (alpha - beta)^2 = 4
anonymous
  • anonymous
Alright, understood then what's the next thing that you did?
BAdhi
  • BAdhi
As per your information you know how to take the sum and product of the roots from the coefficients of the quadratic equation and simply that's what I've done
anonymous
  • anonymous
yea that's the first steps of the working.
anonymous
  • anonymous
Now the third step should be?
anonymous
  • anonymous
Substituting values to previous given identity, 4=(−b9)2−4(79)
anonymous
  • anonymous
alright using the (alpha - beta) square = ... formula
anonymous
  • anonymous
understood.

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