Question

find the root of the equation 1/x+4-1/x-7=11/30,x≠-4,7​

Answer 1

Required Answer:-

Question:

• Find the roots of the given equation.

Solution:

We have,

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

$\sf \implies \dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)}= \dfrac{11}{30}$

$\sf \implies \dfrac{ \cancel{x} - 7- \cancel{x}- 4}{(x + 4)(x - 7)}= \dfrac{11}{30}$

$\sf \implies \dfrac{ - \cancel{ 11}}{(x + 4)(x - 7)}= \dfrac{ \cancel{11}}{30}$

$\sf \implies \dfrac{ -1}{(x + 4)(x - 7)}= \dfrac{1}{30}$

On transposing, we get,

$\sf \implies (x + 4)(x - 7) = -30$

$\sf \implies x(x - 7) + 4(x - 7) = -30$

$\sf \implies {x}^{2} - 7x + 4x - 28 = -30$

$\sf \implies {x}^{2} -3x - 28 = -30$

$\sf \implies {x}^{2} -3x - 28 + 30 = 0$

$\sf \implies {x}^{2} -3x + 2= 0$

Now, the given quadratic equation is in standard form. Split -3 into two parts such that sum of those two parts is -3 and their product is 2.

We found that,

• (-1) × (-2) = 2
• -1 - 2 = -3

So,

$\sf \implies {x}^{2} -x - 2x+ 2= 0$

$\sf \implies x(x -1) - 2(x - 1) = 0$

$\sf \implies (x - 2)(x - 1) = 0$

By zero product rule, either (x - 2) = 0 or (x - 1) = 0,

$\sf \implies x = 1,2$

★ Hence, the roots of the given quadratic equation are 1 and 2.

Answer:

• The roots of the given quadratic equation are 1 and 2.