
How do you solve \[{x^2} - 9x = 0\] using the quadratic formula?
Answer
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Hint: If we have a polynomial of degree ‘n’ then we have ‘n’ number of roots or factors. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step-by-step solution:
Given, \[{x^2} - 9x = 0\].
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We get, \[a = 1\], \[b = - 9\] and \[c = 0\].
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
\[ \Rightarrow x = \dfrac{{ - ( - 9) \pm \sqrt {{{( - 9)}^2} - 4(1)(0)} }}{{2(1)}}\]
\[ \Rightarrow x = \dfrac{{9 \pm \sqrt {81} }}{2}\]
(We know that the product of two negative numbers gives us a positive number.)
We know 81 is a perfect square we have,
\[ \Rightarrow x = \dfrac{{9 \pm 9}}{2}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{9 + 9}}{2}\] and \[x = \dfrac{{9 - 9}}{2}\]
\[ \Rightarrow x = \dfrac{{18}}{2}\] and \[x = \dfrac{0}{2}\]
\[ \Rightarrow x = 9\] and \[x = 0\]. This is the required result.
Thus the values of x are x=9 and x=0.
Additional information:
In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts
Note: We can also solve this directly by taking ‘x’ common. That is
\[\Rightarrow {x^2} - 9x = 0\]
\[\Rightarrow x(x - 9) = 0\]
Then from zero product principle we have
\[\Rightarrow x = 0\] and \[x - 9 = 0\]
\[ \Rightarrow x = 0\] and \[x = 9\]. In both the cases we have the same answer.
Complete step-by-step solution:
Given, \[{x^2} - 9x = 0\].
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We get, \[a = 1\], \[b = - 9\] and \[c = 0\].
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
\[ \Rightarrow x = \dfrac{{ - ( - 9) \pm \sqrt {{{( - 9)}^2} - 4(1)(0)} }}{{2(1)}}\]
\[ \Rightarrow x = \dfrac{{9 \pm \sqrt {81} }}{2}\]
(We know that the product of two negative numbers gives us a positive number.)
We know 81 is a perfect square we have,
\[ \Rightarrow x = \dfrac{{9 \pm 9}}{2}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{9 + 9}}{2}\] and \[x = \dfrac{{9 - 9}}{2}\]
\[ \Rightarrow x = \dfrac{{18}}{2}\] and \[x = \dfrac{0}{2}\]
\[ \Rightarrow x = 9\] and \[x = 0\]. This is the required result.
Thus the values of x are x=9 and x=0.
Additional information:
In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts
Note: We can also solve this directly by taking ‘x’ common. That is
\[\Rightarrow {x^2} - 9x = 0\]
\[\Rightarrow x(x - 9) = 0\]
Then from zero product principle we have
\[\Rightarrow x = 0\] and \[x - 9 = 0\]
\[ \Rightarrow x = 0\] and \[x = 9\]. In both the cases we have the same answer.
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