Question

Solve the following equations by using quadratic formula:
A). \[\dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{4}{x+4};x\ne -1,-2,-4\]
B). \[\dfrac{1}{2x-3}+\dfrac{1}{x-5}=1\dfrac{1}{9};x\ne \dfrac{3}{2},5\]
C). \[{{x}^{2}}+\left( \dfrac{a}{a+b}+\dfrac{a+b}{a} \right)x+1=0\]
D). \[\dfrac{x+3}{x+2}=\dfrac{3x-7}{2x-3};x\ne -2,\dfrac{3}{2}\]

Answer 1

Hint: Take least common multiple on left hand side. Now you get an equation where 2 fractions are present on both sides. Just apply cross multiplication, now you get an equation with both sides as a combination of variables. Now try to make the right hand side as zero. Now you get a quadratic equation of form \[a{{x}^{2}}+bx+c=0\]. By applying quadratic formula roots of this equation are \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step-by-step solution -
(a)Given equation in terms of x in this part, is given by:
\[\Rightarrow \dfrac{1}{x+1}+\dfrac{2}{x+2}=\dfrac{4}{x+4}\]
By taking least common multiple on left hand side, we get,
\[\Rightarrow \dfrac{x+2+2\left( x+1 \right)}{\left( x+1 \right)\left( x+2 \right)}=\dfrac{4}{x+4}\]
By simplifying the above equation, we get it as: -
\[\Rightarrow \dfrac{3x+4}{{{x}^{2}}+3x+2}=\dfrac{4}{x+4}\]
By cross multiplication, we get the equation into form of:
\[\Rightarrow \left( 3x+4 \right)\left( x+4 \right)=4\left( {{x}^{2}}+3x+2 \right)\]
By subtracting the term \[\left( 4{{x}^{2}}+12x+8 \right)\] on both sides, we get,
\[\Rightarrow 3{{x}^{2}}+16x+16-4{{x}^{2}}-12x-8=0\]
By simplifying this equation, we get the equations as:
\[\Rightarrow -{{x}^{2}}+4x+8=0\]
By multiplying “-1” on both sides of equations, we get it as,
\[\Rightarrow {{x}^{2}}-4x-8=0\]
By comparing it to \[a{{x}^{2}}+bx+c=0\], we get a = 1, b = -4, c = -8.
By substituting into formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we get it as,
\[\Rightarrow x=\dfrac{4\pm \sqrt{16-4\left( -8 \right)}}{2}=\dfrac{4\pm \sqrt{16+32}}{2}\]
By simplifying the above, we can write value of x as:
\[\Rightarrow x=2\pm 2\sqrt{3}\]
(b) Given equation in terms of x in this part, is written as:
\[\dfrac{1}{2x-3}+\dfrac{1}{x-5}=1\dfrac{1}{9}\]
By taking least common multiple of left hand side, we get:
\[\Rightarrow \dfrac{x-5+2x-3}{\left( 2x-3 \right)\left( x-5 \right)}=\dfrac{10}{9}\]
By taking cross multiplication, we get it as written below: -
\[\Rightarrow 27x-72=20{{x}^{2}}-130x+150\]
By subtracting the term 27x – 72 on both sides, we get it as:
\[\Rightarrow 20{{x}^{2}}-157x+222=0\]
By using formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], a = 20, b = - 157, c = 222.
\[x=\dfrac{157\pm \sqrt{{{157}^{2}}-4\left( 20 \right)\left( 222 \right)}}{40}=\dfrac{157\pm \sqrt{6889}}{40}\]
So, the roots of given equation are, given as:
\[x=\dfrac{157\pm \sqrt{6889}}{40}=\dfrac{157\pm 83}{40}=\dfrac{74}{40},\dfrac{240}{40}\]
By simplifying the values we get x = 1.85, 6.
(c) Give equation in terms of x in the question is given by:
\[{{x}^{2}}+\left( \dfrac{a}{a+b}+\dfrac{a+b}{a} \right)x+1=0\]
By taking least common multiple, we get it as given:
\[\begin{align}
  & \Rightarrow \left( {{a}^{2}}+ab \right){{x}^{2}}+\left( {{a}^{2}}+{{a}^{2}}+{{b}^{2}}+2ab \right)x+ab+{{b}^{2}}=0 \\
 & \Rightarrow a\left( a+b \right){{x}^{2}}+\left( {{a}^{2}}+{{\left( a+b \right)}^{2}} \right)x+a\left( a+b \right)=0 \\
\end{align}\]
By substituting in the formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[\Rightarrow x=\dfrac{-\left( {{a}^{2}}+{{\left( a+b \right)}^{2}} \right)\pm \sqrt{{{\left( {{a}^{2}}+{{\left( a+b \right)}^{2}} \right)}^{2}}-4{{a}^{2}}{{\left( a+b \right)}^{2}}}}{2a\left( a+b \right)}\]
By simplifying the root, we get it as:
\[\Rightarrow x=\dfrac{-\left( {{a}^{2}}+{{\left( a+b \right)}^{2}} \right)\pm \sqrt{{{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{\left( a+b \right)}^{2}} \right)}^{2}}-2{{a}^{2}}{{\left( a+b \right)}^{2}}}}{2a\left( a+b \right)}\]
By simplifying the term inside the root, we get it as:
\[\begin{align}
  & x=\dfrac{-\left( {{a}^{2}}+{{\left( a+b \right)}^{2}} \right)\pm \left( {{a}^{2}}-{{\left( a+b \right)}^{2}} \right)}{2a\left( a+b \right)} \\
 & x=\dfrac{-2{{\left( a+b \right)}^{2}}}{2a\left( a+b \right)},\dfrac{-2{{a}^{2}}}{2a\left( a+b \right)} \\
\end{align}\]
By simplifying the terms given by x, we get it as:
\[x=\dfrac{-\left( a+b \right)}{a},\dfrac{-a}{\left( a+b \right)}\]
(d) Given equation in terms of x in the question is given by:
\[\Rightarrow \dfrac{x+3}{x+2}=\dfrac{3x-7}{2x-3}\]
As it is given \[x\ne -2,\dfrac{3}{2}\], we know \[x+2\ne 0,2x-3\ne 0\].
By applying cross multiplication on above equation, we get it as:
\[\Rightarrow \left( x+3 \right)\left( 2x-3 \right)=\left( x+2 \right)\left( 3x-7 \right)\]
By simplifying the product terms on both sides, we get it as:
\[\Rightarrow 2{{x}^{2}}-3x+6x-9=3{{x}^{2}}-7x+6x-14\]
By subtracting \[2{{x}^{2}}\] on both sides, we get it as:
\[\Rightarrow {{x}^{2}}-x-14=3x-9\]
By subtracting (3x – 9) on both sides, we get it as:
\[\Rightarrow {{x}^{2}}-4x-5=0\]
By comparing it to \[a{{x}^{2}}+bx+c=0\], we get a = 1, b = -4, c = -5.
By substituting these into the equation of \[n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we get,
\[\Rightarrow x=\dfrac{4\pm \sqrt{16-\left( -20 \right)}}{2}=\dfrac{4\pm \sqrt{36}}{2}\]
By simplifying the root term, we get it as:
\[\Rightarrow x=\dfrac{4\pm 6}{2}\]
By separating the terms, we can write values of x as:
\[\Rightarrow x=\dfrac{10}{2},\dfrac{-2}{2}=5,-1\].

Note: While applying formula take care of negative signs while comparing. As we have –b in the formula a sign change may eventually affect the whole answer. While removing brackets be careful you must multiply the term with constant also generally students multiply to variable and forget about constants.