One root of the quadratic equation (2x-3) (x+5)=0 is -5, then the other root is
Answers
Step-by-step explanation:
2 roots :
Alpha = -5 , Beta
(2x-3) (x+5) = 0
2x²+10x-3x-15 = 0
2x²+7x-15 = 0
Therefore, a = 2, b = 7, c = -15
We know that,
Alpha * Beta = c/a
-5 * Beta = -15/2
Beta = -15/2*-5
Beta = -15/-10 i.e Beta = 15/10
Therefore, Beta = 1.5 or 3/2
The other root is 1.5
Given :
- The quadratic equation is, (2x-3) (x+5)=0
- One root is -5
To find : The other root.
Solution :
We can simply solve this mathematical problem by using the following mathematical process. (our goal is to calculate the other root of the given quadratic equation)
Let, the other root = y
Product of two roots = (-5) × y = -5y (i)
Now,
(2x-3) (x+5) = 0
or, 2x²-3x+10x-15 = 0
or, 2x²+7x-15 = 0
So, the quadratic equation is :
2x²+7x-15 = 0
Comparing with, ax²+bx+c = 0, we get :
- a = 2
- b = 7
- c = -15
We know that,
Product of two roots = c ÷ a
By, putting the available values, we get :
Product of two roots = (-15) ÷ 2 = -7.5 (ii)
By comparing (i) and (ii), we get :
-5y = -7.5
5y = 7.5
y = 7.5/5
y = 1.5
Hence, the other root is 1.5