Find either of the values of x 2x^2+3x-2=3

Mathematics

1 answer:

Serjik [45]3 years ago

8 0

Answer:

x = 1 or x = -5/2

Step-by-step explanation:

Solve for x over the real numbers:

2 x^2 + 3 x - 2 = 3

Divide both sides by 2:

x^2 + (3 x)/2 - 1 = 3/2

Add 1 to both sides:

x^2 + (3 x)/2 = 5/2

Add 9/16 to both sides:

x^2 + (3 x)/2 + 9/16 = 49/16

Write the left hand side as a square:

(x + 3/4)^2 = 49/16

Take the square root of both sides:

x + 3/4 = 7/4 or x + 3/4 = -7/4

Subtract 3/4 from both sides:

x = 1 or x + 3/4 = -7/4

Subtract 3/4 from both sides:

Answer: x = 1 or x = -5/2

You might be interested in

How do extraneous solutions arise from radical equations?

Blababa [14]

In general, extraneous solutions arise<span> when we perform non-invertible operations on both sides of an </span>equation<span>. (That is, they sometimes </span>arise, but not always.) ... Solvingequations<span> involving square roots involves squaring both sides of an</span>equation<span>. I hope that this helps you out, Have a wonderful day!!</span>

5 0

3 years ago

Read 2 more answers

2/8 + 6/12 what is that eqaul to

Papessa [141]

$\frac{2}{8} + \frac{6}{12} \\ \\ \frac{1}{4} + \frac{1}{2} \\ \\ LCD = 4 \\ \\ \frac{1}{4} + \frac{2}{4} \\ \\ \frac{1+2}{4} \\ \\ \frac{3}{4} \\ \\ Answer: \fbox {3/4} \ or \ \fbox {0.75}$