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1 month ago

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.

Mathematics

1 answer:

lawyer [12.5K]1 month ago

6 0

Answer:

$x=-2.38$

$x=-4.62$

Step-by-step explanation:

The question is $(2x+3)^2+8(2x+3)+11=0$

We let $u=2x+3$ , so the equation becomes:

$u^2+8u+11=0$

Where $a=1, b=8, c=11$

By applying the quadratic formula, we arrive at:

Quadratic formula: $\frac{-b+-\sqrt{b^2-4ac} }{2a}$

Substituting yields: $\frac{-8+-\sqrt{(8)^2-4(1)(11)} }{2(1)}\\=\frac{-8+-\sqrt{20} }{2}\\=\frac{-8+-2\sqrt{5} }{2}\\=-4+\sqrt{5}, -4-\sqrt{5} }$

We let $u=2x+3$ , thus x calculates to:

$u=2x+3\\(-4+\sqrt{5})=2x+3\\x=\frac{-7+\sqrt{5}}{2}=-2.38$

and

$u=2x+3\\(-4-\sqrt{5})=2x+3\\x=\frac{-7-\sqrt{5}}{2}=-4.62$

The solutions to the equation are $x=-2.38$ (rounded to 2 decimal places) and $x=-4.62$ (rounded to 2 decimal places)

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