Solve the following quadratic equation for all values of xx in simplest form. 3(x+1)²-15=-12
1 answer:
You might be interested in
If the third term of the aritmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Given the third term of the aritmetic sequence is 126 and sixty fourth term is 3725.
We are required to find the first term of the arithmetic sequence.
Arithmetic sequence is a series in which all the terms have equal difference.
Nth term of an AP=a+(n-1)d
=a+(3-1)d
126=a+2d--------1
=a+(64-1)d
3725=a+63d------2
Subtract second equation from first equation.
a+2d-a-63d=126-3725
-61d=-3599
d=59
Put the value of d in 1 to get the value of a.
a+2d=126
a+2*59=126
a+118=126
a=126-118
a=8
=a+(1-1)d
=8+0*59
=8
Hence if the third term of the arithmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Learn more about arithmetic progression at brainly.com/question/6561461
#SPJ1
Answer:
Option D)Neither solution is extraneous.
Step-by-step explanation:
we have
we know that
two possible solutions are x=-7 and x=1
<u><em>Verify each solution</em></u>
Substitute each value of x in the expression above and interpret the results
1) For x=-7
----> is true
therefore
x=-7 is not a an extraneous solution
2) For x=1
----> is true
therefore
x=1 is not a an extraneous solution
therefore
Neither solution is extraneous