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+3x−5=0
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+ab+ac=0 . Next we add and subtract a term such that we get a perfect square of the form (a+b)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+3x−5=0
Now the given equation is a quadratic equation of the form ax2+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+23x−25=0
Now adding and subtracting 169 on LHS we get
⟹x2+23x+169−169−25=0
⟹x2+23x+169−1649=0
Now writing the above equation in the form a2+2ab+b2 and taking the constant to RHS we get,
x2+2×43×1+(43)2=1649
Now we know that the expansion of (a+b)2 is a2+2ab+b2
Hence we can write x2+2(43)(1)+(43)2 as (x+43)2
Therefore we have (x+43)2=1649
Taking square root on both sides we get,
(x+43)=±47
Now we have two solutions,
Either x+43=47 or x+43=−47
Taking the terms on RHS we get,
x=47−43 or x=−47−43
Hence the values of x are x = 1 or x=−410=−25
Hence the solution of the given quadratic equation is x = 1 or x=−25 .
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 ab by 2 and square. Hence the term which needs to be added and subtracted is (2ab)2.