Question

solve the following pair of linear equations using quadratic formula 1/x+4-1/x-7=11/30,x=-4,7​

Answer 1

GIVEN :-

$\\ \sf \: \dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$

TO FIND :-

• Roots of the polynomial.

SOLUTION :-

We have ,

$\\ \implies\sf \: \dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30} \\ \\ \implies\sf \: \dfrac{x - 7 - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \\ \\ \implies\sf \: \dfrac{ \cancel x - 7 - \cancel x - 4}{x(x - 7) + 4(x - 7)} = \dfrac{11}{30} \\ \\ \implies\sf \: \dfrac{ - 11}{ {x}^{2} - 7x + 4x - 28} = \dfrac{11}{30} \\ \\ \implies\sf \: \dfrac{ - 11}{ {x}^{2} - 3x - 28 } = \dfrac{11}{30}$

$\\ \implies\sf \: - 11(30) = 11( {x}^{2} - 3x - 28) \\ \\ \implies\sf \: {x}^{2} - 3x - 28 = \dfrac{ - \cancel{11}(30)}{ \cancel{11}} \\ \\ \implies\sf \: {x}^{2} - 3x - 28 = - 30 \\ \\ \implies\sf \: {x}^{2} - 3x - 28 + 30 = 0 \\ \\ \implies\underline{ \sf \: {x}^{2} - 3x + 2 = 0}$

Now , the equation is in the form of ax² + bx + c.

Here ,

• a = 1
• b = -3
• c = 2

QUADRATIC FORMULA :-

$\\ \clubsuit \boxed{\sf \: roots = \dfrac{ - b ±\sqrt{ {b}^{2} - 4ac } }{2a} }$

Putting values,

$\\ \longmapsto\sf \: roots = \dfrac{ - ( - 3) ± \sqrt{ {( - 3)}^{2} - 4(1)(2) } }{2(1)} \\ \\ \longmapsto\sf \: roots = \dfrac{3± \sqrt{9 - 8} }{2} \\ \\ \longmapsto\sf \: roots = \dfrac{3± \sqrt{1} }{2}$

$\longmapsto\sf \: roots = \dfrac{3 + 1}{2} \: , \: \dfrac{3 - 1}{2} \\ \\ \longmapsto\sf \: roots = \dfrac{4}{2} \: , \: \dfrac{2}{2} \\ \\ \longmapsto\underbrace{\boxed{\sf \: roots = 2 \: , \: 1}}$

Hence , roots are 2,1.