Answer:
(a). none
(b). x=4, x=12
Explanation:
(a). (x+3)2+4=0
x2+9+6x+4=0
x2+6x+13=0 -------------------------------------------------- (1)
Comparing this equation with standard equation, ax2+bx+c=0, we have,
a=1
b=6
c=13
For the equation to have real solutions, b2-4ac>0.
We have,
b2=6^2=36
4ac=4*1*13=52
So, b2-4ac=36-52=-16
so, b2-4ac<0.
Hence, the equation does not have a real solution.
(b).
Squaring on both sides
2x+1=(7-x)2
2x+1=49+x2-14x
x2-14x+49-2x-1=0
x2-16x+48=0
x2-12x-4x+48=0
x(x-12)-4(x-12)=0
(x-12)(x-4)=0
So we have,
(x-12)=0 => x=12
(x-4)=0 =>x=4
So, real solutions are x=4, x=12.
f (x) = e.13, oo) k. none x-3 Find the real solution(s) for the following equations....
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