Answer:
Option 3 [tex]\frac{67}{441}[/tex]
Step-by-step explanation:
step 1
Find the roots of the quadratic equation
we have
[tex]3x^{2}+5x-7=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2}+5x-7=0[/tex]
so
[tex]a=3\\b=5\\c=-7[/tex]
substitute in the formula
[tex]x=\frac{-5(+/-)\sqrt{5^{2}-4(3)(-7)}} {2(3)}[/tex]
[tex]x=\frac{-5(+/-)\sqrt{109}} {6}[/tex]
[tex]x=\frac{-5+\sqrt{109}} {6}[/tex]
[tex]x=\frac{-5-\sqrt{109}} {6}[/tex]
step 2
Let
[tex]\alpha=\frac{-5+\sqrt{109}} {6}[/tex]
[tex]\beta=\frac{-5-\sqrt{109}} {6}[/tex]
we need to calculate
[tex]\frac{1}{(3\alpha+5)^{2}}+ \frac{1}{(3\beta+5)^{2}}[/tex]
step 3
Calculate [tex](3\alpha+5)^{2}[/tex]
[tex](3\alpha+5)^{2}=[3(\frac{-5+\sqrt{109}} {6})+5]^{2}[/tex]
[tex]=[(\frac{-5+\sqrt{109}} {2})+5]^{2}[/tex]
[tex]=[(\frac{-5+\sqrt{109}+10} {2})]^{2}[/tex]
[tex]=[(\frac{5+\sqrt{109}} {2})]^{2}[/tex]
[tex]=[(\frac{25+10\sqrt{109}+109} {4})][/tex]
[tex]=[(\frac{134+10\sqrt{109}} {4})][/tex]
[tex]=[(\frac{67+5\sqrt{109}} {2})][/tex]
step 4
Calculate [tex](3\beta+5)^{2}[/tex]
[tex](3\beta+5)^{2}=[3(\frac{-5-\sqrt{109}} {6})+5]^{2}[/tex]
[tex]=[(\frac{-5-\sqrt{109}} {2})+5]^{2}[/tex]
[tex]=[(\frac{-5-\sqrt{109}+10} {2})]^{2}[/tex]
[tex]=[(\frac{5-\sqrt{109}} {2})]^{2}[/tex]
[tex]=[(\frac{25-10\sqrt{109}+109} {4})][/tex]
[tex]=[(\frac{134-10\sqrt{109}} {4})][/tex]
[tex]=[(\frac{67-5\sqrt{109}} {2})][/tex]
step 5
substitute
[tex]\frac{1}{(3\alpha+5)^{2}}+ \frac{1}{(3\beta+5)^{2}}[/tex]
[tex]\frac{1}{[(\frac{67+5\sqrt{109}} {2})]}+ \frac{1}{[(\frac{67-5\sqrt{109}} {2})]}[/tex]
[tex]\frac{2}{67+5\sqrt{109}} +\frac{2}{67-5\sqrt{109}}\\ \\\frac{2(67-5\sqrt{109})+2(67+5\sqrt{109})}{(67+5\sqrt{109})(67-5\sqrt{109})} \\ \\\frac{268}{1764}[/tex]
Simplify
Divide by 4 both numerator and denominator
[tex]\frac{268}{1764}=\frac{67}{441}[/tex]
