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:

& \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{aligned}$$ By substituting in the formula, $$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{aligned} & 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{aligned}$$ By simplifying the terms given by x, we get it as: $$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.