Question

Solve using quadratic formula:
1) $ {{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0 $
2) $ 9{{x}^{2}}-9\left( a+b \right)x+\left( 2{{a}^{2}}+5a+2{{b}^{2}} \right) $

Answer 1

Hint: First write the equation and compare it to the general quadratic equation, by comparing find values of a, b, c and substitute them into a quadratic formula. Simplify this formula to get the value of x. This value of x is the required solution.

Complete step-by-step answer:
Given equation in the question in form of
 $ {{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0............\left( 1 \right) $
General quadratic equation is written in the form of:
 $ {{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0..................\left( 2 \right) $
Roots of equation (2) are given by formula of x:
 $ x=\dfrac{-{{b}_{1}}\pm \sqrt{{{b}_{1}}^{2}-4{{a}_{1}}{{c}_{1}}}}{2{{a}_{1}}} $
By comparing we can say value of a, b, c as:
 $ {{a}_{1}}={{p}^{2}},{{b}_{1}}={{p}^{2}}-{{q}^{2}},{{c}_{1}}=-{{q}^{2}} $
By substituting these, we can write it in form of
 $ x=\dfrac{-\left( {{p}^{2}}-{{q}^{2}} \right)\pm \sqrt{{{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}}-4\left( {{p}^{2}} \right)\left( -{{q}^{2}} \right)}}{2{{p}^{2}}} $
We know the formula of square given by the equation:
 $ {{\left( a-b \right)}^{2}}+4ab={{\left( a+b \right)}^{2}} $
By substituting this we can write the term inside square root as
 $ x=\dfrac{-\left( {{p}^{2}}-{{q}^{2}} \right)\pm \sqrt{{{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}}}{2{{p}^{2}}} $
By simplifying above term, we can get value of x as
 $ x=\dfrac{-{{p}^{2}}+{{q}^{2}}\pm \left( {{p}^{2}}+{{q}^{2}} \right)}{2{{p}^{2}}} $
By separating the roots, we can write values of x as
 $ x=\dfrac{-{{p}^{2}}+{{q}^{2}}+{{p}^{2}}+{{q}^{2}}}{2{{p}^{2}}},\dfrac{-{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-{{q}^{2}}}{2{{p}^{2}}} $
By simplifying both of the above, we can write them as
 $ x=\dfrac{2{{q}^{2}}}{2{{p}^{2}}},\dfrac{-2{{p}^{2}}}{2{{p}^{2}}} $
By simplifying, we can write values of x in the form of
 $ x=\dfrac{{{q}^{2}}}{{{p}^{2}}},-1 $
Given equation in the question, part (ii) we can write it as
 $ 9{{x}^{2}}-9\left( a+b \right)x+\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right) $
By comparing it to equation (2), we can say their values
 $ {{a}_{1}}=9,{{b}_{1}}=-9\left( a+b \right),{{c}_{1}}=2{{a}^{2}}+5ab+2{{b}^{2}} $
By substituting these into formula we can write it as
 $ x=\dfrac{+\left( 9\left( a+b \right) \right)\pm \sqrt{{{\left( 9\left( a+b \right) \right)}^{2}}-4\left( 9 \right)\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)}}{18} $
Taking square root term separately, we can write it in form of
 $ \sqrt{81{{\left( a+b \right)}^{2}}-36\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)} $
By simplifying the term with “-“ we get it in form:
 $ \sqrt{81{{\left( a+b \right)}^{2}}-72{{a}^{2}}-180ab-72{{b}^{2}}} $
By simplifying the square term, we substitute equation:
 $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $
We get it as:
\[\sqrt{81\left( {{a}^{2}}+2ab+{{b}^{2}} \right)-72{{a}^{2}}-72{{b}^{2}}-180ab}\]
By simplifying above term, we can write it as:
 $ \sqrt{81{{a}^{2}}+81{{b}^{2}}+162ab-72{{a}^{2}}-72{{b}^{2}}-180ab} $
By simplifying above term, we can write it as:
 $ \sqrt{9{{a}^{2}}+9{{b}^{2}}-18ab} $
By taking 9 as common inside, we get it in the form of:
 $ \sqrt{9\left( {{a}^{2}}+{{b}^{2}}-2ab \right)} $
By using the algebraic identity of $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ , we get:
\[\sqrt{9{{\left( a-b \right)}^{2}}}=3\left( a-b \right)\]
By substituting this into equation of x, we get it as
 $ x=\dfrac{9\left( a+b \right)\pm 3\left( a-b \right)}{18}=\dfrac{9a+9b+3a-3b}{18},\dfrac{9a+9b-3a+3b}{18} $
By simplifying, we can write in the form of:
 $ x=\dfrac{a+2b}{3},\dfrac{2a+b}{3} $
Therefore the values of x all roots for the respective equation.

Note: While applying quadratic formula take care of sign –b in the starting. Generally students confuse this step. It is better to solve the square root separately for more clarity. While taking values of x separately take care while taking the “-“ in $ \pm $ symbol. Because students make a lot of mistakes in “-“ they forget to subtract the root itself. So be careful.