Question

Select the correct answer.

Answer 1

Answer:

Option C. The given system has two solutions.

Solution:

The given equations are,

$y = -x + 1$

$y = -x^2 + 4x-2$

From the equation we can say,

$-x+1 = -x^2+ 4x-2$

$\Rightarrow-x^{2}+4 x-2+x-1=0$

$\Rightarrow-x^{2}+5 x-3=0$

We know that the quadratic formula to solve this,

x has two values which are $\frac{(-b+\sqrt{b^{2}-4 a c})}{2 a} \ and \ \frac{(-b-\sqrt{\left.b^{2}-4 a c\right)}}{2 a}$

Here, a = (-1), b = 5 , c = -3

So, $x=\frac{(-5+\sqrt{(5)^{2}-4 x(-1)} \times(-3))}{2 \times(-1)}=\frac{(-5+\sqrt{25-12})}{-2}=\frac{(-5+\sqrt{13})}{-2}=\frac{(5-\sqrt{13})}{2}$

Again $x=\frac{(5+\sqrt{13})}{2}$

Hence, x has two solutions.