# If the roots of equation x^2 + qx + p=0 are each 1 less than the roots of the equation x^2 + px + q = 0 then (p+q) is equal to?

Apr 1, 2018

given that the roots of equation ${x}^{2} + q x + p = 0$ are each 1 less than the roots of the equation ${x}^{2} + p x + q = 0$.

Let the roots of the equation ${x}^{2} + p x + q = 0$ be $\alpha \mathmr{and} \beta$. So the roots of equation ${x}^{2} + q x + p = 0$ will be $\left(\alpha - 1\right) \mathmr{and} \left(\beta - 1\right)$

So we can write

$\alpha + \beta = - p \mathmr{and} \alpha \beta = q$

again

$\alpha + \beta - 2 = - q \mathmr{and} \left(\alpha - 1\right) \left(\beta - 1\right) = p$

Comparing we get

$\alpha + \beta - 2 = - \alpha \beta \ldots \ldots \ldots \left[1\right]$

and

$\left(\alpha - 1\right) \left(\beta - 1\right) = - \alpha - \beta$

$\implies \alpha \beta - \alpha - \beta + 1 = - \alpha - \beta$

$\implies \alpha \beta = - 1 \ldots . . \left[2\right]$

By {1] and [2]

$\alpha + \beta - 2 = - \left(- 1\right) = 1$

$\alpha + \beta = 3$

Now from first two relations

$q - \left(- p\right) = \alpha \beta - \left(\alpha + \beta\right) = - 1 - 3 = - 4$

$\implies p + q = - 4$

Apr 1, 2018

$p + q = - 4$

#### Explanation:

Assuming

${x}^{2} + q x + p = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = 0$
${x}^{2} + p x + q = \left(x - {x}_{1} + 1\right) \left(x - {x}_{2} + 1\right) = 0$

and comparing coefficients we get

$\left\{\begin{matrix}p = - \left({x}_{1} + {x}_{2}\right) \\ q = {x}_{1} {x}_{2} \\ p = 1 - \left({x}_{1} + {x}_{2}\right) + {x}_{1} {x}_{2} \\ q = 2 - \left({x}_{1} + {x}_{2}\right)\end{matrix}\right.$

now calling ${y}_{1} = {x}_{1} + {x}_{2}$ and ${y}_{2} = {x}_{1} {x}_{2}$ we have

$\left\{\begin{matrix}p = - {y}_{1} \\ q = {y}_{2} \\ p = 1 - {y}_{1} + {y}_{2} \\ q = 2 - {y}_{1}\end{matrix}\right.$

Solving

$p = - 3 , q = - 1 , {y}_{1} = 3 , {y}_{2} = - 1$

hence

$p + q = - 4$