find the points of intersection of the parabolas Y=(1r2 x^2) and Y=10x - 2.
To find the points of intersection of the parabolas \(y = x^2\) and \(y = 10x - 2\), we can set the two equations equal to each other and solve for \(x\).
Given:
\(y = x^2\) -----(1)
\(y = 10x - 2\) -----(2)
Setting equation (1) equal to equation (2), we have:
\(x^2 = 10x - 2\)
To solve this quadratic equation, let's rearrange it to standard form:
\(x^2 - 10x + 2 = 0\)
To find the values of \(x\), we can apply the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
From equation (1), we can see that \(a = 1\), \(b = -10\), and \(c = 2\).
Substituting these values into the quadratic formula, we get:
\(x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(2)}}{2(1)}\)
Simplifying further:
\(x = \frac{10 \pm \sqrt{100 - 8}}{2}\)
\(x = \frac{10 \pm \sqrt{92}}{2}\)
\(x = \frac{10 \pm 2\sqrt{23}}{2}\)
Now, we have two possible values for \(x\). Let's calculate the corresponding \(y\) values using equation (2):
For \(x = \frac{10 + 2\sqrt{23}}{2}\):
\(y = 10\left(\frac{10 + 2\sqrt{23}}{2}\right) - 2\)
\(y = 5(10+2\sqrt{23}) - 2\)
\(y = 50 + 10\sqrt{23} - 2\)
\(y = 48 + 10\sqrt{23}\)
For \(x = \frac{10 - 2\sqrt{23}}{2}\):
\(y = 10\left(\frac{10 - 2\sqrt{23}}{2}\right) - 2\)
\(y = 5(10-2\sqrt{23}) - 2\)
\(y = 50 - 10\sqrt{23} - 2\)
\(y = 48 - 10\sqrt{23}\)
Therefore, the points of intersection of the parabolas \(y = x^2\) and \(y = 10x - 2\) are:
\(\left(\frac{10 + 2\sqrt{23}}{2}, 48 + 10\sqrt{23}\right)\)
\(\left(\frac{10 - 2\sqrt{23}}{2}, 48 - 10\sqrt{23}\right)\)