can you solve the equations. write the solutions as integers if possible otherwise as radical expressions?

Question

(this means squared ^2)
x^2=25 , y^2=81 , 3a^2=147 , x^2=121 , m^2=0 , k^2=49 , y^2=36 , a^2=20 , t^2=63 , -3x^2=27 , x^2+4=16 , 2b^2-7=-7 , 16-x^2=12 , x^2-16=0 , 2x^2-37=35 , 3-x^2=50 , 8-2x^2=-33 , 3x^2-58=50, 5x^2+20=4 3x^2-8=21 9X^2+15=40 3X^2-8=21 2X^2-8=33 5x^2-10=3 Good Luck and thank you for the effort.

Answer 1

WAY too much work for one question. You should post this as several separate questions next time, maybe 4-5 problems per question.

Here are the last four answers:

9X^2+15=40
9x^2 = 25
x^2 = 25/9
x = 5/3

3X^2-8=21
3x^2 = 29
x^2 = 29/3
x = sqrt(29)/sqrt(3)
x = sqrt(87)/3

2X^2-8=33
2x^2 = 41
x^2 = 41/2
x = sqrt(41)/sqrt(2)
x = sqrt(82)/2

5x^2-10=3
5x^2 = 13
x^2 = 13/5
x = sqrt(13)/sqrt(5)
x = sqrt(65)/5

Note that I placed all the answers in standard form without a radical in the denominator. For example the last one: sqrt(13)/sqrt(5) ... multiply by sqrt(5)/sqrt(5) to remote the radical from the denominator.

Answer 2

a million. 16-x^(2)=12 Reorder the polynomial 16-x^(2) alphabetically from left to good, beginning with the utmost order time period. -x^(2)+16=12 considering that 16 does no longer contain the variable to bathe up for, pass it to the right-hand area of the equation by subtracting 16 from both area. -x^(2)=-16+12 upload 12 to -16 to get -4. -x^(2)=-4 Multiply each and every time period contained in the equation by -a million. -x^(2)*-a million=-4*-a million Multiply -x^(2) by -a million to get x^(2). x^(2)=-4*-a million Multiply -4 by -a million to get 4. x^(2)=4 Take the sq. root of both area of the equation to get rid of the exponent on the left-hand area. x=+ o or - sq. root of (4) Pull all acceptable sq. roots out from below the novel. therefore, remove both because that's a acceptable sq.. x= = or - 2 First, replace contained in the + component of the to locate the first answer. x=2 next, replace contained in the - component of the to locate the second one answer. x=-2 the completed answer is the effect of both the + and - parts of the answer. x= 2, -2 2. a^(2)=20 a=(20) a=+ o or - 2= sq. root of (5) a=2square root of (5) a=-2square root of (5) a=2square root of (5),-2square root of (5) a=4.40 seven,-4.40 seven 3. x^(2)+4=16 x^(2)=-4+16 x^(2)=12 x==+ o or - sq. root of(12) x==+ o or -2 sq. root of(3) x=2 sq. root of(3) x=-2 sq. root of (3) x=2square root of(3),-2square root of(3) x= 3.40 six,-3.40 six 4. 8-2x^2=-33+3x^2-8=21 x=2,-2 8-2x^2=-33-3x^2-8=21 x=2,-2

Answer 3

Ok do your own math homework. I'll get you started. To get rid of a squared, take the square root of both sides. That is the last thing you do, first move all constants do the other side and divide to get the variable squared by itself.