Four times the square of a number is 45 more than eight times the number. What is the number?

Question

this is a algerbraic question . please help one of my problems for homework!!!!!!!

Answer 1

4x^2 = 8x + 45

4x^2 - 8x - 45 = 0

(2x + 5) (2x - 9) = 0

x = -2.5 (-5/2) or 4.5 (9/2)

or you can use the quadratic to solve 4x^2 - 8x - 45= 0

Good luck.

Answer 2

4x^ is 45 more than 8x
therefore 4x^ is more than 8x, equation will be
4x^ - 45 = 8x
4x^ - 8x - 45 = 0
(2x - 9)(2x + 5) = 0
2x - 9 = 0 and 2x + 5 = 0
x = 9/2 and x = - 5/2

Answer 3

call the extensive type n. you pick to study 4 cases the sq. of the extensive type (this is 4n^2) to eight cases the extensive type (this is 8n) on condition that 4n^2 is forty 5 greater desirable than 8n you could say that 4n^2=8n+forty 5 or 4n^2-8n-forty 5=0 As it incredibly is a quadratic equation it could have 2 suggestions and fixing for them calls for the quadratic formulation, for an equation interior certainly one of those ax^2+bx+c=0 x=[-b+/-sqrt(b^2 - 4ac)]/2a So right here we've a=4, b=-8 and c=-forty 5 yet we are fixing for n no longer x. n=[8+/-sqrt(sixty 4-4(4)(-forty 5))/2(4) n=[8+/-sqrt(sixty 4+720)]8 n=[8+/-28]/8 n1=(8+28)/8=4.5 n2=(8-28)/8=-2.5 to study them we merely placed those numbers lower back into the unique equation 4n^2=8n+forty 5 while n=4.5 4n^2=4(4.5^2)=eighty one 8n=36 eighty one-36=forty 5 so 4n^2 is forty 5 greater desirable than 8n. while n=-2.5 4n^2=4(-2.5^2)=25 8n=-20 25-(-20)=forty 5 so lower back 4n^2 is forty 5 greater desirable than 8n so those are the suggestions.

Answer 4

4x^2 = 8x + 45
4x^2 - 8x - 45 = 0
(2x + 5) (2x -9) = 0
x = -5/2 or 9/2

Answer 5

4 n^2 = 8n + 45
4 n^2 - 8n - 45 = 0
(2n + 5)(2n - 9) = 0
2n = -5
n = -5/2
2n = 9
n = 9/2