the equation [tex]10 {r}^{2} + 13r - 3 = 0[/tex]has solution of the form [tex]r = \frac{n + \sqrt{d} } {m} [/tex]use the quadratic formula to solve this equation and find the appropriate integer values of N, M and D. Do not worry about simplifying the D N=____ ; D=_____ M=_____



Answer :

10r^2 + 13r - 3 = 0

[tex]\begin{gathered} \frac{-b\text{ }\pm\sqrt[]{b^2\text{ - 4ac}}}{2a} \\ \frac{-13\text{ }\pm\sqrt[]{13^2\text{ - 4(10)(-3)}}}{2(10)} \\ \frac{-13\text{ }\pm\sqrt[]{169\text{ +120}}}{20} \\ \frac{-13\text{ }\pm\sqrt[]{289}}{20} \\ \end{gathered}[/tex]

a = 10

b= 13

c = -3

a) From the result, we conclude that

N = -13 D = 289 M = 20

b)

[tex]\begin{gathered} \frac{-13\text{ }\pm17}{20} \\ r1\text{ = }\frac{-13\text{ + 17}}{20}\text{ r2 = }\frac{-13\text{ - 17}}{20} \\ r1\text{ = }\frac{4}{20}\text{ r2 = }\frac{-30}{20} \\ r1\text{ = }\frac{1}{5}\text{ r2 =-}\frac{15}{10\text{ }}=-\frac{3}{2} \\ \end{gathered}[/tex]

r1 = 1/5 r2 = -3/2