Question

Please help...Use the quadratic and a calculator to solve the following equation. Round intermediate calculations and final answers to two decimal places. 6.25y^2 + 12y + 5.76 = 0
i got y = .96, -96 but it was wrong should i round up to 1, -1

Answer 1

if you require the answer correct to dec places to do not iround intermediate results to 2 dp's

Answer 2

right so I dont understand the statement Round intermediate calculations and final answers to two decimal places

Answer 3

2 dec places i meant to say

Answer 4

Is it bcuz it is saying Use the quadratic and a calculator to solve

Answer 5

i make the solutios -0.96

Answer 6

there is only one solution

Answer 7

so you think incluce the 0

Answer 8

can a solution set have only one thing in it?

Answer 9

yes

Answer 10

6.25y^2 + 12y + 5.76 = 0

Rationalize the coefficients.
\[\frac{25}{4}y^2+12y+\frac{144}{25}=0\]Multiply each side by 100 and simplify.\[625 y^2+1200 y+576=0 \]Use the binomial theorem or, to pull a rabbit out of the hat:\[625 y^2+1200 y+576=(25 y+24)^2 \]\[y=\frac{24}{25}\text{ }\text{or}\text{ }0.96 \]

Answer 11

if you graph this function it will just touch the x axis at -0.96

Answer 12

I stand corrected. Yes, the sign is negative. Thank you.

Answer 13

ok

Answer 14

so you are saying that the only solution in the solution set is -0.96??

Answer 15

There are two solutions, each with identical values.

Answer 16

so both would be -0.96

Answer 17

yes

Answer 18

but I probably only need to enter it once?

Answer 19

the books always say two identical solutions but to be honest i dont understand why

Answer 20

Woohoo it was correct...thank you both sooooo much!!!

Answer 21

I use Mathematica to do the calculations.\[\text{Solve}\left[(25 y+24)^2\text{==}0,y\right]=\left\{\left\{y\to -\frac{24}{25}\right\},\left\{y\to -\frac{24}{25}\right\}\right\} \]Notice that it returned two values for y. I would put down as an answer, x = -0.96 twice.

Does this answer you questions?

Answer 22

no probs popcorn

Answer 23

Your welcome.