Question

Find the possible values of b for which the equation 9x^2 + bx + 7 = o has roots which differ by 2.

Answer 1

This time round i hope someone can post the sum of roots and product of roots method *please*

Answer 2

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 $$\Delta=b^2-4ac$$

Answer 3

I have not learnt modulus yet

Answer 4

going to soon

Answer 5

if you want only from sum and products of the roots,
use,
$$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$$

Answer 6

Can you show me that steps please?

Answer 7

to derive out the possible values of b

Answer 8

Yours sincerely, very much appreciate it.

Answer 9

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

Answer 10

how did you get (alpha - beta) square ?

Answer 11

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

Answer 12

Alright, understood then what's the next thing that you did?

Answer 13

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

Answer 14

yea that's the first steps of the working.

Answer 15

Now the third step should be?

Answer 16

Substituting values to previous given identity,
4=(−b9)2−4(79)

Answer 17

alright using the (alpha - beta) square = ... formula

Answer 18

understood.