Question

Find the roots of the quadratic equation 9(p + q)2
x
2 + 6(p + q)x + 1 = 0 .

Answer 1

AnswEr :

The roots of the given equation are:

$\bf -\dfrac{1}{3(p+q)} \: \: and \: \: -\dfrac{1}{3(p+q)}$

SoluTion :

Given , the quadratic equation,

9(p + q)² x² + 6(p + q)x + 1 = 0

$\sf \implies9(p+q)^{2}x^{2} + 6(p+q)x + 1 =0\\\\ \sf \implies\{3(p+q) x \} ^{2}+2\times \{3(p+q)x \}\times1+1^{2}=0 \\\\ \sf\implies \{3(p+q)x + 1 \}^{2}= 0 \\\\ \sf\implies \{3(p+q)x + 1 \} \{3(p+q)x+1\} = 0$

Thus the roots are :

$\sf 3(p+q)x+1=0 \: \: and \: \: 3(p+q)x+1=0\\\\ \sf \dashrightarrow x = -\dfrac{1}{3(p+q)} \: \: and \: \: x = -\dfrac{1}{3(p+q)}$