Question #445f7

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Question #445f7

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Answer 1 · 2018-02-14T08:18:59

$5 x^{2} - 10 x - 2 = 0$ $5x^2-10x-2=0$

Explanation:

If the equation is $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ then alpha+beta= b/a and alpha.beta=c/a.
Comparing $a x^{2} + b x + c = 0, x^{2} + (- 2) x + 5 = 0$ $ax^2+bx+c=0, x^2+(-2)x+5=0$ we get alpha+beta = -2 and alpha.beta=5.
Now, 1/alpha + 1/beta =(alpha+beta)/alpha.beta= -2/5.
Hence new equation is:
$x^{2} + (- 2) x + (- \frac{2}{5}) = 0$ $x^2+(-2)x+(-2/5)=0$
Or, $5 x^{2} - 10 x - 2 = 0$ $5x^2-10x-2=0$

Answer 2 · 2018-02-14T10:48:49

see a solution process below;

Explanation:

Let $A$ $A$ and $B$ $B$ be alpha and beta respectively..

$x^{2} - 2 x + 5 = x^{2} - (A + B) x + A B$ $x^2-2x + 5 = x^2 -(A+B)x + AB$

$A + B = 2$ $A + B = 2$

$A B = 5$ $AB = 5$

When $A + B$ $A + B$ and $\frac{1}{A} + \frac{1}{B}$ $1/A+1/B$ are the roots of the equation,

The expression becomes;

$x^{2} - (A + B + (\frac{1}{A} + \frac{1}{B}) x + [(A + B) (\frac{1}{A} + \frac{1}{B})]$ $x^2-(A+B+(1/A+1/B)x+[(A+B)(1/A+1/B)]$

Therefore;

$x^{2} - (A + B + (\frac{A + B}{A B}) x + [(A + B) (\frac{A + B}{A B})]$ $x^2-(A+B+((A+B)/(AB))x+[(A+B)((A+B)/(AB))]$

Substituting the values;

$x^{2} - (2 + \frac{2}{5}) x + (2 \cdot \frac{2}{5})$ $x^2-(2+2/5)x+(2*2/5)$

$x^{2} - (\frac{12}{5}) x + \frac{4}{5}$ $x^2-(12/5)x+4/5$

Multiply through by $5$ $5$

$5 x^{2} - 12 x + 4$ $5x^2 - 12x + 4$

Hope this helps!

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