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Question

Question: Solve the given equation by complete square method \(2{{x}^{2}}+3x-5=0\)...

Solve the given equation by complete square method 2x2+3x5=02{{x}^{2}}+3x-5=0

Explanation

Solution

To solve the given quadratic equation by complete square method first we will divide the whole equation by a. Now we have an equation of the form x2+ba+ca=0{{x}^{2}}+\dfrac{b}{a}+\dfrac{c}{a}=0 . Next we add and subtract a term such that we get a perfect square of the form (a+b)2{{\left( a+b \right)}^{2}} on LHS . Now taking the constant on RHS and then taking square root we will get the required value of x.

Complete step by step answer:
Now consider the given equation 2x2+3x5=02{{x}^{2}}+3x-5=0
Now the given equation is a quadratic equation of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0 where a = 2, b = 3 and c = - 5.
Now let us divide the whole equation by a which is 2. Hence, we get
x2+32x52=0{{x}^{2}}+\dfrac{3}{2}x-\dfrac{5}{2}=0
Now adding and subtracting 916\dfrac{9}{16} on LHS we get
    x2+32x+91691652=0\implies {{x}^{2}}+\dfrac{3}{2}x+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{5}{2}=0
    x2+32x+9164916=0\implies {{x}^{2}}+\dfrac{3}{2}x+\dfrac{9}{16}-\dfrac{49}{16}=0
Now writing the above equation in the form a2+2ab+b2{{a}^{2}}+2ab+{{b}^{2}} and taking the constant to RHS we get,
x2+2×34×1+(34)2=4916{{x}^{2}}+2\times \dfrac{3}{4}\times 1+{{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{49}{16}
Now we know that the expansion of (a+b)2{{\left( a+b \right)}^{2}} is a2+2ab+b2{{a}^{2}}+2ab+{{b}^{2}}
Hence we can write x2+2(34)(1)+(34)2{{x}^{2}}+2\left( \dfrac{3}{4} \right)\left( 1 \right)+{{\left( \dfrac{3}{4} \right)}^{2}} as (x+34)2{{\left( x+\dfrac{3}{4} \right)}^{2}}
Therefore we have (x+34)2=4916{{\left( x+\dfrac{3}{4} \right)}^{2}}=\dfrac{49}{16}
Taking square root on both sides we get,
(x+34)=±74\left( x+\dfrac{3}{4} \right)=\pm \dfrac{7}{4}
Now we have two solutions,
Either x+34=74x+\dfrac{3}{4}=\dfrac{7}{4} or x+34=74x+\dfrac{3}{4}=-\dfrac{7}{4}
Taking the terms on RHS we get,
x=7434x=\dfrac{7}{4}-\dfrac{3}{4} or x=7434x=-\dfrac{7}{4}-\dfrac{3}{4}
Hence the values of x are x = 1 or x=104=52x=-\dfrac{10}{4}=-\dfrac{5}{2}
Hence the solution of the given quadratic equation is x = 1 or x=52x=-\dfrac{5}{2} .

Note: Now note that when finding the term which is supposed to be added and subtracted to the equation to form a complete square divide the middle term which is ba\dfrac{b}{a} by 2 and square. Hence the term which needs to be added and subtracted is (b2a)2{{\left( \dfrac{b}{2a} \right)}^{2}}.