Quadratic equation

**mgmanoj** · August 1st 2012, 04:14 PM

Let r and s be the roots of the quadratic x²+ bx + c where b and c are constant, if (r-1)(s-1) = 7 find b + c

**Prove It** · August 1st 2012, 08:31 PM

Originally Posted by mgmanoj

Let r and s be the roots of the quadratic x²+ bx + c where b and c are constant, if (r-1)(s-1) = 7 find b + c

$\displaystyle \begin{align*} x^2 + b\,x + c &= 0 \\ x^2 + b\,x + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c &= 0 \\ \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4} + \frac{4c}{4} &= 0 \\ \left(x + \frac{b}{2}\right)^2 + \frac{4c - b^2}{4} &= 0 \\ \left(x + \frac{b}{2}\right)^2 &= \frac{b^2 - 4c}{4} \\ x + \frac{b}{2} &= \frac{\pm \sqrt{b^2 - 4c}}{2} \\ x &= \frac{-b \pm \sqrt{b^2 - 4c}}{2} \end{align*}$

So we can say $\displaystyle \begin{align*} r = \frac{-b - \sqrt{b^2 - 4c}}{2} \end{align*}$ and $\displaystyle \begin{align*} s = \frac{-b + \sqrt{b^2 - 4c}}{2} \end{align*}$ . Since we know $\displaystyle \begin{align*} (r - 1)(s - 1) = 7 \end{align*}$ , we can say

$\displaystyle \begin{align*} \left(\frac{-b - \sqrt{b^2 - 4c}}{2} - 1\right)\left(\frac{-b + \sqrt{b^2 - 4c}}{2} - 1\right) &= 7 \\ \left(\frac{-b - 2 - \sqrt{b^2 - 4c}}{2}\right)\left(\frac{-b - 2 + \sqrt{b^2 - 4c}}{2}\right) &= 7 \\ \frac{\left(- b - 2 - \sqrt{b^2 - 4c} \right) \left( - b - 2 + \sqrt{b^2 - 4c} \right) }{4} &= 7 \\ \left(-b-2 - \sqrt{b^2 - 4c}\right)\left(-b-2 + \sqrt{b^2 - 4c}\right) &= 28 \\ \left(- b - 2\right)^2 - \left(b^2 - 4c \right) &= 28 \\ b^2 + 4b + 4 - b^2 + 4c &= 28 \\ 4b + 4c &= 24 \\ 4(b + c) &= 24 \\ b + c &= 6 \end{align*}$

**mgmanoj** · August 2nd 2012, 04:11 AM

Thanks - I was half way down to the problem - but did not substitute - this confirms my theory. Thanks again.

**Soroban** · August 2nd 2012, 09:14 AM

Hello, mgmanoj!

$\text{Let }r\text{ and }s\text{ be the roots of the quadratic: }\: x^2+\,bx\,+\,c\,\text{ where }b\text{ and }c\text{ are constants.}$
$\text{If }\,(r-1)(s-1) \:=\: 7,\,\text{ find }b + c.$

$\text{Since }r\text{ and }s\text{ are roots: }\:\begin{Bmatrix}r + s &=& \text{-}b \\ rs &=& c \end{Bmatrix} \;\;{\color{blue}[1]}$

$\text{Given: }\:(r-1)(s-1) \:=\:7 \quad\Rightarrow\quad rs - r - s + 1 \:=\:7 \quad\Rightarrow\quad rs - (r+s) \:=\:6$

$\text{Substitute }{\color{blue}[1]}\!:\;c - (\text{-}b) \:=\:6 \quad\Rightarrow\quad b +c \:=\:6$

**richard1234** · August 2nd 2012, 09:19 AM

I second Soroban's solution...always helps to look at the sum and product of the roots of the quadratic.

**Wilmer** · August 2nd 2012, 04:26 PM

Originally Posted by richard1234

I second Soroban's solution...always helps to look at the sum and product of the roots of the quadratic.

I third it !

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