Question

How do you solve \[{y^2} - 90 = 13y\]?

Answer 1

Hint: After rearranging the given equation we will have a polynomial equation. The obtained polynomial is of degree 2. Instead of ‘x’ as a variable we have ‘y’ as a variable. We can solve this by using factorization methods or by using quadratic formulas. We use quadratic formula if factorization fails. We know that a polynomial equation has exactly as many roots as its degree.

Complete step by step answer:
Given,
\[{y^2} - 90 = 13y\]
Rearranging we have,
\[{y^2} - 13y - 90 = 0\]
On comparing the given equation with the standard quadratic equation \[A{y^2} + By + C = 0\]. Where ‘A’ and ‘B’ are coefficients of \[{y^2}\] and coefficient of ‘y’ respectively.
We have \[A = 1\], \[B = - 13\] and \[C = - 90\].
The standard form of the factorization of quadratic equation is \[A{y^2} + {B_1}y + {B_2}y + C = 0\], which satisfies the condition \[{B_1} \times {B_2} = A \times C\] and \[{B_1} + {B_2} = B\].
We can write the given equation as \[{y^2} - 18y + 5y - 90 = 0\], where \[{B_1} = - 18\] and \[{B_2} = 5\]. Also \[{B_1} \times {B_2} = ( - 18) \times (5) = - 90(A \times C)\] and \[{B_1} + {B_2} = ( - 18) + (5) = - 13(B)\].
\[{y^2} - 13y - 90 = 0 \Rightarrow {y^2} - 18y + 5y - 90 = 0\]
\[{y^2} - 18y + 5y - 90 = 0\]
In the first two terms we take ‘y’ as common and in the remaining term we take 5 as common,
\[y(y - 18) + 5(y - 18) = 0\]
Again taking \[(y - 18)\] as common we have,
\[(y - 18)(y + 5) = 0\]
By zero multiplication property we have,
\[y - 18 = 0\] and \[y + 5 = 0\].

\[ \Rightarrow y = 18\] and \[y = - 5\]. These are the roots of the given problem.

Note: 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.