# Question a0eac

Apr 1, 2017

$a = 1 , b = - 5 n , c = 4 {n}^{2}$, where $n$ = the number

#### Explanation:

Let $n =$ the number, let $4 n =$ 4 times the number

$\left(x - n\right) \left(x - 4 n\right) = 0$

Distribute: ${x}^{2} - 5 n x + 4 {n}^{2} = 0$

Compare to $a {x}^{2} + b x + c = 0$:

$a = 1 , b = - 5 n , c = 4 {n}^{2}$

Check:
Let $n = 5 , 4 x = 20$
$\left(x - 5\right) \left(x - 20\right) = 0$
${x}^{2} - 25 x + 100 = 0$
$a = 1 , b = - 5 \cdot 5 = - 25 , c = 4 \cdot {5}^{2} = 100$

Let $n = - 5 , 4 x = - 20$
$\left(x + 5\right) \left(x + 20\right) = 0$
${x}^{2} + 25 x + 100 = 0$
$a = 1 , b = 5 \cdot 5 = 25 , c = 4 \cdot {\left(- 5\right)}^{2} = 100$

Apr 1, 2017

$4 {b}^{2} = 25 a c .$

#### Explanation:

Let $\alpha \mathmr{and} \beta$ be the Roots of the given Quadr.

By what is given, we may, assume, $\beta = 4 \alpha \ldots . . \left(1\right) .$

We also know that, alpha+beta=-b/a....(2), &, alpha*beta=c/a...(3).

(1) & (2) rArr 5alpha=-b/a, i.e., alpha=-b/(5a)....(4).

(1) & (3) rArr 4alpha^2=c/a, i.e., alpha^2=c/(4a)......(5).

"Finally, "(4) & (5) rArr c/(4a)={-b/(5a)}^2=b^2/(25a^2).#

$\therefore 25 a c = 4 {b}^{2} ,$ is the desired relation btwn. $a , b , c .$

Enjoy Maths.!