Question

please solve this in quadratic equations​

Answer 1

In algebra, a quadratic equation is any equation that can be rearranged in standard form as ax^{2}+bx+c=0 where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no ax^2 term

.Formula

ax^2+bx+c=0

a, b, c = known numbers, where a ≠ 0

x = the unknown

Answer 2

Correct Question :

$\displaystyle \text{{Solve for x : }}Solve for x :$

$\displaystyle \sf{{\frac{1}{a+b+x} =\frac{1}{a} +\frac{1}{b} +\frac{1}{x} }}$

Answer:

$\displaystyle \sf{{x=-a \ \ OR \ \ x=-b}}x=−a OR x=−b$

Step-by-step explanation:

$\displaystyle \sf{{Given : }}Given:$

$\displaystyle \sf{{\frac{1}{a+b+x} =\frac{1}{a} +\frac{1}{b} +\frac{1}{x} }}$

$</p><p>\displaystyle \text{{Take \frac{1}{\text{x}} L.H.S. side : }}$

$\begin{gathered}\displaystyle \sf{{\frac{1}{a+b+x}-\frac{1}{x} =\frac{1}{a} +\frac{1}{b} }}\\\\\\\displaystyle \sf{{\frac{x-(a+b+x)}{(a+b+x)(x)}=\frac{b+a}{ab} }}\end{gathered}$

$</p><p>\begin{gathered}\displaystyle \sf{{\frac{x-a-b-x}{(a+b+x)(x)}=\frac{a+b}{ab} }}\\\\\\\displaystyle \sf{{\frac{-(a+b)}{(ax+bx+x^2)}=\frac{a+b}{ab} }}\end{gathered}$

$</p><p>\displaystyle \text{{Now (a + b) get cancel out : }}Now (a + b)$

$</p><p>\begin{gathered}\displaystyle \sf{{\frac{-1}{(ax+bx+x^2)}=\frac{1}{ab} }}\\\\\\\displaystyle \sf{{ax+bx+x^2=-ab}}\\\\\\\displaystyle \sf{{x^2+ax+bx+ab=0}}\end{gathered}$

$\displaystyle \text{{Take out 'x' and 'b' common : }}$

$\begin{gathered}\displaystyle \sf{{x(x+a)+b(x+a)=0}}\\\\\\\displaystyle \sf{{(x+a)(x+b)}}\end{gathered} </p><p>$

$\begin{gathered}\displaystyle \sf{{\rightarrow x+a=0}}\\\\\\\displaystyle \sf{{\rightarrow x = -a}}\\\\\\\displaystyle \sf{{OR}}\\\\\\\displaystyle \sf{{\rightarrow x+b=0}}\\\\\\\displaystyle \sf{{\rightarrow x=-b}}\end{gathered} </p><p>$

$\displaystyle \text{{Therefore , the value of x is '-a' OR ' -b' .}}$