Which point is a solution to this linear quadratic system? $y = x^{2} + 4 x + 3$ and $y = 2 x + 6$

(- 3, 0)

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(1, - 8)

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(3, 0)

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(0, - 3)

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Solution

The correct option is B $(- 3, 0)$
$y = x^{2} + 4 x + 3$ -------(i)
$y = 2 x + 6$ ---------------(ii)
Putting the value of y from (ii) in (i), we get
$x^{2} + 4 x + 3 = 2 x + 6$
$=> x^{2} + 2 x - 3 = 0$
$=> x^{2} + 3 x - x - 3 = 0$
$=> x (x + 3) - (x + 3) = 0$
$=> (x + 3) (x - 1) = 0$
$=> x = - 3$ or $x = 1$
$=> y = 2 (- 3) + 6$ or $y = 2 (1) + 6$
$=> y = 0$ or $y = 8$
Points are $(- 3, 0)$ and $(1, 8)$
Thus $(- 3, 0)$ is a solution.

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Which point is a solution to this linear quadratic system? y=x2+4x+3 and y=2x+6

Which point is a solution to this linear quadratic system? $y = x^{2} + 4 x + 3$ and $y = 2 x + 6$