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# help on algebra

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Suppose that there is only one value of $$x$$ for which the distance from $$(5,6)$$ to $$(3x-7,ax+2)$$ is $$4$$. If a $$a \neq 0,$$ what is $$a$$?

Nov 27, 2020

#1
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Suppose that there is only one value of x  for which the distance from $$P_1(5,6)$$  to $$P_2(3x-7,ax+2)$$  is 4 . If  $$a \neq 0,$$ what is  a?

Hello Guest!

$$d^2=(y_2-y_1)^2+(x_2-x_1)^2$$

$$4^2=(ax+2-6)^2+(3x-7-5)^2\\ 16=(ax-4)^2+(3x-12)^2\\ 16=a^2x^2-8ax+16+9x^2-72x+144\\ f(x)=(a^2+9)x^2-(8a+72)x+144=0$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

$$x = {(8a+72) \pm \sqrt{{\color{blue} (8a+72)^2-4\cdot (a^2+9)\cdot 144}} \over 2\cdot (a^2+9)}$$

When the radical is zero, the function f (x) has a single solution.

$$(8a+72)^2-4\cdot (a^2+9)\cdot 144=0$$      | $$a \neq 0$$

$$a=2.25$$

$$x = 3.2$$

!

Nov 28, 2020
edited by asinus  Nov 28, 2020
edited by asinus  Nov 28, 2020
edited by asinus  Nov 28, 2020
#2
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Thank you!

Guest Nov 28, 2020