Question #a0eac

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Question #a0eac

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Answer 1 · 2017-04-01T17:16:17

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

Explanation:

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

$(x - n) (x - 4 n) = 0$ $(x - n)( x - 4n) = 0$

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

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

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

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

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

Answer 2 · 2017-04-01T18:09:22

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

Explanation:

Let $α and β$ $alpha and beta$ be the Roots of the given Quadr.

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

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

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

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

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

$:. 25ac=4b^2,$ is the desired relation btwn. $a,b,c.$

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