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How do you solve for x in $y = {x^2} - 2x$ ?

seo-qna
Last updated date: 22nd Jul 2024
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Answer
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Hint: In this question, we are given a quadratic equation in terms of x as x is raised to the power 2 and we have to solve for the value of x. A quadratic equation is solved by using factorization or the quadratic formula or completing the square method. We will first convert the given equation to the standard form, that is, $a{x^2} + bx + c = 0$ and then compare them to find the value of the coefficients a, b and c. Then we will put these values in the quadratic formula and solve it. This way we will get the value of x in terms of y. As the degree of the equation is 2, so we will get 2 values of x.

Complete step-by-step solution:
We are given that $y = {x^2} - 2x$
Rearranging the equation, we get –
${x^2} - 2x - y = 0$
On comparing it with $a{x^2} + bx + c = 0$ , we get –
$a = 1,\,b = - 2\,and\,c = - y$
The quadratic formula is given as –
$
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4 \times 1 \times ( - y)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 4y} }}{2} = \dfrac{{2 \pm 2\sqrt {1 + y} }}{2} \\
   \Rightarrow x = 1 \pm \sqrt {1 + y} \\
 $
Hence when $y = {x^2} - 2x$ , we get $x = 1 + \sqrt {1 + y} $ or $x = 1 - \sqrt {1 + y} $ .

Note: When the unknown variable quantity in an algebraic expression is raised to some non-negative integer as the power, we get a polynomial equation, and the highest power of the unknown quantity in the equation is known as the degree of the polynomial equation. The highest exponent of x in the given equation is 2, so it has a degree 2 and the polynomial equation is a quadratic equation. We can verify that the answer obtained is correct by putting the obtained value of x in the given equation.