Answer:
The Quadratic Polynomial is
2 x² +x -4=0
Using the Determinant method to find the roots of this equation
[tex]x=\frac{-1\pm \sqrt{1^2-4*2*(-4)}}{2\times 2}\\\\ x=\frac{-1 \pm \sqrt{32}}{4}{\text{Using the formula}}x=\frac{-b\pm\sqrt{D}}{2a}, {\text{where}},D=b^2-4 ac[/tex]
For, the Quadratic equation , ax²+ b x+c=0
(b) x²+x=0
x × (x+1)=0
x=0 ∧ x+1=0
x=0 ∧ x= -1
You can look the problem in other way
the two Quadratic polynomials are
2 x²+x-4=0, ∧ x²+x=0
x²= -x
So, 2 x²+x-4=0,
→ -2 x+x-4=0
→ -x -4=0
→x= -4
∨
x² +x² +x-4=0
x²+0-4=0→→x²+x=0
→x²=4
x=√4
x=2 ∧ x=-2
As, you will put these values into the equation, you will find that these values does not satisfy both the equations.
So, there is no solution.
You can solve these two equation graphically also.
Answer:
1/2 and -2 are the first answers to the question
1/16 and 1/16 are the next answers
1/4 and 33/16 are the last ones
and for the multiple choice answer which is last is A
Step-by-step explanation:
