Question

1. 2x² + 11x + 14=0 2. 2y2 + 13y + 21 = 0​

Answer 1

Answer:

1. x = -2 and $x = -\dfrac{7}{2}$

2. y = -3 and $y = -\dfrac{7}{2}$

Step-by-step explanation:

The given equations are-

$2x^{2} + 11x + 14 = 0$          ----------- 1

$2y^{2} + 13y + 21 = 0$         ----------- 2

Solving equation 1-

We have-

$2x^{2} + 11x + 14 = 0$

$2x^{2} + (7+4)x + 14 = 0$

$2x^{2} + 7x + 4x + 14 = 0$

$x(2x+7) + 2(2x+7) = 0$

$(2x+7)(x+2) = 0$

If (x + 2) = 0

Then x = -2

If (2x + 7) = 0

Then 2x = -7

$x = -\dfrac{7}{2}$

Solving equation 2

$2y^{2}+13y + 21 = 0$

$2y^{2} + (7+6)y + 21 = 0$

$2y^{2} + 7y + 6y + 21 = 0$

$y(2y + 7) + 3(2y+7) = 0$

$(2y+7)(y+3) = 0$

If (y + 3) = 0

Then, y = -3

If (2y + 7) = 0

Then, 2y = -7

$y = -\dfrac{7}{2}$