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#### Monoxdifly

##### Well-known member
Aug 6, 2015
312
The equation $$\displaystyle (a-1)x^2-4ax+4a+7=0$$ with a is a whole number has positive roots. If $$\displaystyle x_1>x_2$$ then $$\displaystyle x_2-x_1=...$$
A. –8
B. –5
C. –2
D. 2
E. 8

Since the equation has positive roots then $$\displaystyle x_1>0$$ and $$\displaystyle x_2>0$$ thus $$\displaystyle x_1+x_2>0$$ and $$\displaystyle x_1x_2>0$$

$$\displaystyle x_1+x_2>0$$
$$\displaystyle \frac{-(-4a)}{a-1}>0$$

$$\displaystyle x_1x_2>0$$
$$\displaystyle \frac{4a+7}{a-1}>0$$

However I progressed, I couldn't determine a as a single value and only found it as a set of certain whole numbers. Can you help me to find the single value of a? Once I know that. I guess I can continue on my own.

#### skeeter

##### Well-known member
MHB Math Helper
Mar 1, 2012
980
$b^2-4ac = 16a^2 - 4(a-1)(4a+7) > 0 \implies 4(7-3a) > 0 \implies a \in \{0,1,2\}$

of those three possible values for $a$, only $a=2$ yields positive roots ... $x = 3$ & $x = 5$
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