Question

Complete the square to solve \[{{x}^{2}}-2x-35=0\]?

Answer 1

Hint: As we know in the given numerical here we have to complete the square of the question, so here we have to use the technique of completing square i. e in desirable form as \[{{\text{a}}^{\text{2}}}\text{+ }{{\text{b}}^{\text{2}}}\] means both the variable should be completed square.

Complete step by step solution:
In this question we have to do the squaring of variables so writing the equation are as below
\[{{x}^{2}}-2x-35=0\] Further we have to expand it in the form as \[{{\text{a}}^{\text{2}}}\text{+ }{{\text{b}}^{\text{2}}}\] that it will be in the form of complete square.
Thus the give equation we have as follow:
\[{{x}^{2}}-2x-35=0\]
Now we have to make it in the form as \[{{\text{a}}^{\text{2}}}\text{+ }{{\text{b}}^{\text{2}}}\] so we have to evaluate it,
Thus to make it in above form we are adding one that will unbalance the equation so we are subtracting the equation with one we will get it as,
\[{{x}^{2}}-2x+1-1-35=0\]
So in the above equation we have added and subtracted with one.
Further solving we are taking a likely term one side and another term on the other side we get it as.
We can write \[{{x}^{2}}-2x+1\] as \[{{\left( x-1 \right)}^{2}}\]thus we had written in the term of \[{{\left( \text{a - b} \right)}^{\text{2}}}\text{= }{{\text{a}}^{\text{2}}}\text{ - 2ab + }{{\text{b}}^{\text{2}}}\]
No also subtracting minus one with minus thirty-five we get as thirty six, so we can write it as,
\[{{\left( x-1 \right)}^{2}}-36\]
Thus we had obtained the equation above by solving.
Further solving the above equation we can write it as
\[\left( \left( x-1 \right)-6 \right)\left( \left( x-1 \right)+6 \right)\]
 So above we had again simplify the equation,
Further splitting in the squaring from we get,
\[\left( x-7 \right)\left( x+5 \right)\]
Thus we get the two value of \[x\] we obtained \[7\] and \[-5\]
\[\text{x = 7}\] or \[\text{x = -5}\]
Hence by completing the square we obtained the value of \[7\] and \[-5\].

Note: While completing the square we have to evaluate and add or subtract the value to make it a perfect square. Make terms one side and other terms the other side.