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Question: How do you solve \({{x}^{2}}-2x=99\)?...

How do you solve x2−2x=99{{x}^{2}}-2x=99?

Explanation

Solution

We first add 1 to both sides of the equation x2−2x=99{{x}^{2}}-2x=99. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we add 1 to the both sides to find the value of xx for x2−2x=99{{x}^{2}}-2x=99.

Complete step by step solution:
We need to find the solution of the given equation x2−2x=99{{x}^{2}}-2x=99.
We first add 1 to both sides of x2−2x=99{{x}^{2}}-2x=99. We use the identity a2−2ab+b2=(a−b)2{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}.
x2−2x+1=99+1 ⇒(x−1)2=100 \begin{aligned} & {{x}^{2}}-2x+1=99+1 \\\ & \Rightarrow {{\left( x-1 \right)}^{2}}=100 \\\ \end{aligned}
We interchanged the numbers for a=x,b=1a=x,b=1.
Now we have a quadratic equation (x−1)2=100{{\left( x-1 \right)}^{2}}=100.
We need to find the solution of the given equation (x−1)2=100{{\left( x-1 \right)}^{2}}=100.
We take square root on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
(x−1)2=100=102 ⇒(x−1)=±10 \begin{aligned} & \sqrt{{{\left( x-1 \right)}^{2}}}=\sqrt{100}=\sqrt{{{10}^{2}}} \\\ & \Rightarrow \left( x-1 \right)=\pm 10 \\\ \end{aligned}
Now we add 1 to the both sides of the equation (x−1)=±10\left( x-1 \right)=\pm 10 to get value for variable xx.
(x−1)+1=±10+1 ⇒x=11,−9 \begin{aligned} & \left( x-1 \right)+1=\pm 10+1 \\\ & \Rightarrow x=11,-9 \\\ \end{aligned}
The given quadratic equation has two solutions and they are x=−9,11x=-9,11.

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation x2−2x=99{{x}^{2}}-2x=99.
The simplified form is x2−2x−99=0{{x}^{2}}-2x-99=0.
We know for a general equation of quadratic ax2+bx+c=0a{{x}^{2}}+bx+c=0, the value of the roots of x will be x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.
In the given equation we have x2−2x−99=0{{x}^{2}}-2x-99=0. The values of a, b, c is 1,−2,−991,-2,-99 respectively.
x=−(−2)±(−2)2−4×1×(−99)2×1=2±4002=2±202=−9,11x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times \left( -99 \right)}}{2\times 1}=\dfrac{2\pm \sqrt{400}}{2}=\dfrac{2\pm 20}{2}=-9,11.
The given quadratic equation has two solutions and they are x=−9,11x=-9,11.