Question

How do you solve \[2{z^2} - 3 = - 5z\] ?

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 ‘z’ 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 solution:
Given,
 \[2{z^2} - 3 = - 5z\]
Rearranging we have,
 \[2{z^2} + 5z - 3 = 0\]
On comparing the given equation with the standard quadratic equation \[A{z^2} + Bz + C = 0\] .
We have \[A = 2\] , \[B = 5\] and \[C = - 3\] .
Here factorization method files. Because the standard form of the factorization of quadratic equation is \[A{z^2} + {B_1}z + {B_2}z + C = 0\] , 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 \[2{z^2} + 6z - 1z - 3 = 0\] , where \[{B_1} = 6\] and \[{B_2} = - 1\] . Also \[{B_1} \times {B_2} = (6) \times ( - 1) = - 6(A \times C)\] and \[{B_1} + {B_2} = (6) + ( - 1) = 5(B)\] .
 \[ \Rightarrow 2{z^2} + 5z - 3 = 2{z^2} + 6z - 1z - 3 = 0\]
 \[2{z^2} + 6z - 1z - 3 = 0\]
In the first two terms we take ‘2z’ as common and in the remaining term we take -1 as common,
 \[2z(z + 3) - 1(z + 3) = 0\]
Again taking \[(z + 3)\] as common we have,
 \[(z + 3)(2z - 1) = 0\]
By zero multiplication property we have,
 \[ \Rightarrow z + 3 = 0\] and \[2z - 1 = 0\] .
 \[ \Rightarrow z = - 3\] and \[2z = 1\]
 \[ \Rightarrow z = - 3\] and \[z = \dfrac{1}{2}\]
These are the roots of the given problem.
So, the correct answer is “ \[ z = - 3\] and \[z = \dfrac{1}{2}\]”.

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.