Question

What are the solutions of the system?

Answer 1

Answer:

Choice C is correct answer.

Step-by-step 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

Answer:

Option C. No solution is the right answer.

Step-by-step 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 $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$=\frac{+2\pm \sqrt{4-4\times 1\times 5}}{2}$

$=\frac{2\pm \sqrt{4-20}}{2}$

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