+0

# Given that the absolute value of the difference of the two roots of $ax^2 + 5x - 3 = 0$ is $\frac{\sqrt{61}}{3}$, and a is positive, what

+2
91
1
+147

Given that the absolute value of the difference of the two roots of $ax^2 + 5x - 3 = 0$ is $\frac{\sqrt{61}}{3}$, and a is positive, what is the value of a?

Feb 24, 2021

#1
+11297
+1

Given that the absolute value of the difference of the two roots of $$ax^2+5x-3=0$$  is $$\sqrt{\frac{61}{3}}$$ , and a is positive, what is the value of a?

Hello powclaire!

$$\dfrac{\sqrt{61}}{3} =\dfrac{-5+ \sqrt{25+12a}}{2a} -\dfrac{-5- \sqrt{25+12a}}{2a}\\ =\dfrac{-5+ \sqrt{25+12a}+5+\sqrt{25+12a}}{2a}$$

$$\dfrac{\sqrt{25+12a}}{a}=\dfrac{\sqrt{61}}{3}\\ \dfrac{25+12a}{a^2}=\dfrac{61}{9}$$

$$61a^2-108a+225=0$$

$$a=3$$

!

Feb 24, 2021
edited by asinus  Feb 24, 2021