Answer :
Answer:
2.)-1,5/2 is your answer
Step-by-step explanation:
2x2 - 3x = 5
2x²-3x-5=0
do ing middle term factorization
2x²-5x+2x-5=0
x(2x-5)+1(2x-5)=0
(2x-5)(x+1)=0
either
x=5/2
or
x=-1
Answer:
[tex] \displaystyle \: 2) - 1,\frac{5}{2} [/tex]
Step-by-step explanation:
we are given a quadratic equation
[tex] \displaystyle \: {2x}^{2} - 3x = 5[/tex]
we have to determine the roots
to do so first bring it to standard form i.e
ax²+bx+c=0
move left hand side expression to right hand side and change its sign:
[tex] \displaystyle \: {2x}^{2} - 3x - 5 =0[/tex]
since we moved left hand side expression to right hand side there's only 0 left to the left hand side
we'll use factoring method to solve the quadratic
first, we need to express the middle term as a sum or subtraction of two different terms
to do so
rewrite -3x as 2x-5x:
[tex] \displaystyle \: {2x}^{2} + 2x - 5x- 5 =0[/tex]
now we have two common factors 2x and -5 so
factor them out:
[tex] \displaystyle \: 2x ({x} + 1 )- 5(x + 1) =0[/tex]
group:
[tex] \displaystyle \: (2x - 5) (x + 1) =0[/tex]
since 2x-5 and x+1 both equal 0
separate the equation as two different equation:
[tex] \displaystyle \: \begin{cases}2x - 5=0 \\ x + 1 = 0\end{cases}[/tex]
let's work with the first equation:
add 5 to both sides:
[tex] \displaystyle \: 2x - 5 + 5 = 0 + 5 \\ 2x = 5[/tex]
divide both sides by 2:
[tex] \displaystyle \: \frac{2x}{2} = \frac{5}{2} \\ \therefore \: x_{} = \frac{5}{2} [/tex]
let's work with the second one
cancel 1 from both sides:
[tex] \displaystyle \: x + 1 - 1 = 1 - 1 \\ \therefore \: x = - 1[/tex]
hence,
our answer choice is [tex]\text{option 2}[/tex]