What is the point of intersection of the straight line 3× - y + 11 = 0 and the parabola y = ×^2 + 4× + 5?

 What is the point of intersection of the straight line 3× - y + 11 = 0 and the parabola y = ×^2 + 4× + 5?💝

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Mathe Solution 👇👇👇👇

To find the point of intersection between the straight line $3𝑥-𝑦+11=0$ and the parabola $𝑦={𝑥}^2+4𝑥+5$, we need to solve these two equations simultaneously.

First, express $𝑦$ from the line equation:

$𝑦=3𝑥+11$

Now, substitute $𝑦=3𝑥+11$ into the parabola equation $𝑦={𝑥}^2+4𝑥+5$:

$3𝑥+11={𝑥}^2+4𝑥+5$

Rearrange this equation to form a standard quadratic equation:

${𝑥}^2+4𝑥+5-3𝑥-11=0$

Simplify the equation:

${𝑥}^2+𝑥-6=0$

Next, solve this quadratic equation using the quadratic formula $𝑥=\frac{-𝑏\pm \sqrt{{𝑏}^2-4𝑎𝑐}}{2𝑎}$, where $𝑎=1$, $𝑏=1$, and $𝑐=-6$:

$𝑥=\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-6)}}{2\cdot 1}$ $𝑥=\frac{-1\pm \sqrt{1+24}}{2}$ $𝑥=\frac{-1\pm \sqrt{25}}{2}$ $𝑥=\frac{-1\pm 5}{2}$

This gives us two solutions for $𝑥$:

$𝑥=\frac{-1+5}{2}=\frac{4}{2}=2$ $𝑥=\frac{-1-5}{2}=\frac{-6}{2}=-3$

Now, we find the corresponding $𝑦$ values for each $𝑥$ by substituting back into $𝑦=3𝑥+11$:

For $𝑥=2$:

$𝑦=3(2)+11=6+11=17$

For $𝑥=-3$:

$𝑦=3(-3)+11=-9+11=2$

Therefore, the points of intersection are:

$(2,17)$ $(-3,2)$

Thus, the points of intersection of the straight line $3𝑥-𝑦+11=0$ and the parabola $𝑦={𝑥}^2+4𝑥+5$ are $(2,17)$ and $(-3,2)$.