Question

What are the solutions of the system?

y = x^2 + 2x +3

y = 4x - 2

a. (-1, -6) and (-3, -14)
b. (-1, 10) and (3, -6)
c. no solution
d. (-1, -6) and (3, 10)

Answer 1

Choice C is correct answer.

Explanation:

Two equations are given:

y = x²+2x+3 eq(1)

y = 4x-2 eq(2)

As both equations are equal to y.So,

x²+2x+3 = 4x-2

Adding -4x and 2 to both sides of above equation,we get

x²+2x+3-4x+2 = 4x-2-4x+2

Adding like terms , we get

x²-2x+5 = 0

ax²+bx+c = 0 is general quadratic equation.

x = (-b±√b²-4ac) / 2a is quadratic formula.

Comparing above equation with general equation,we get

a = 1, b = -2 and c = 5

Putting above value in quadratic formula,we get

x = (-(-2)±√(-2)²-4(1)(5) ) / 2(1)

x = (2±√4-20) / 2

x = (2±√-16) / 2

D = b²-4ac = -16

Hence, D is not real.

There is no solution of given system of equations.

Answer 2

Option C. No solution is the right answer.

Explanation:

Here the given equations are y = x²+2x+3 -----(1)

and y = 4x-2 -------(2)

Now we substitute the value of y from equation 2 into 1.

x²+2x+3 = 4x-2

x²+2x+3-2x = 4x-2-2x

x²+3 = 2x-2

x²+3-2x = 2x-2x-2

x²-2x+3 = -2

x²-2x+5 = 0

Then value of
`x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a}`

`=(+2\pm √(4-4* 1* 5))/(2)`

`=(2\pm √(4-20))/(2)`

Since in this solution √(-20) is not defined. Therefore there is no solution.