Question

find the root of 1/(x+4) -1/(x-7)=11/30 using quadratic formula​

Answer 1

Answer :

The roots are 2 and 1

Step-by-step explanation :

$\sf \dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30} \\\\\\ \dfrac{x-7}{(x+4)(x-7)} -\dfrac{x+4}{(x+4)(x-7)} =\dfrac{11}{30} \\\\\\ \dfrac{x-7-(x+4)}{(x+4)(x-7)} =\dfrac{11}{30} \\\\\\ \dfrac{x-7-x-4}{x(x-7)+4(x-7)} =\dfrac{11}{30} \\\\\\ \dfrac{-\not{11}}{x^2-7x+4x-28}=\dfrac{\not{11}}{30} \\\\\\ -30=x^2-7x+4x-28 \\\\ x^2-3x-28=-30 \\\\ x^2-3x-28+30=0 \\\\ x^2-3x+2=0$

The quadratic equation we got is x² - 3x + 2 = 0

It is of the form ax² + bx + c = 0

a = 1 , b = -3 , c = 2

The quadratic formula is :

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

Substituting a = 1 , b = -3 , c = 2 ;

     $\sf x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(2)} }{2(1)} \\\\ x=\dfrac{3 \pm \sqrt{9-8} }{2} \\\\ x=\dfrac{3 \pm \sqrt{1}}{2} \\\\ x=\dfrac{3 \pm 1}{2} \\\\ x=\dfrac{3+1}{2} ,\dfrac{3-1}{2} \\\\ x=\dfrac{4}{2},\dfrac{2}{2} \\\\ x=2,1$

The roots are 2 and 1

Verification :

• Put x = 2,

$\sf \dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30} \\\\\\ \dfrac{1}{2+4}-\dfrac{1}{2-7}=\dfrac{11}{30} \\\\\\ \dfrac{1}{6}-\dfrac{1}{-5}=\dfrac{11}{30} \\\\\\ \sf \dfrac{1}{6}+\dfrac{1}{5}=\dfrac{11}{30} \\\\\\ \dfrac{5+6}{30}=\dfrac{11}{30} \\\\\\ \dfrac{11}{30} =\dfrac{11}{30} \\\\ LHS=RHS$

• Put x = 1,

$\sf \dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30} \\\\\\ \dfrac{1}{1+4}-\dfrac{1}{1-7}=\dfrac{11}{30} \\\\\\ \dfrac{1}{5}-\dfrac{1}{-6}=\dfrac{11}{30} \\\\\\ \sf \dfrac{1}{5}+\dfrac{1}{6}=\dfrac{11}{30} \\\\\\ \dfrac{6+5}{30}=\dfrac{11}{30} \\\\\\ \dfrac{11}{30} =\dfrac{11}{30} \\\\ LHS=RHS$

Hence verified!