Question

solve

Answer 1

\[(\sqrt{9x+67})^2=(x+5)^2\]\[9x+67= x^2+10x+25\]\[x^2+10x-9x+25-67=0\]\[x^2+x-42=0\]\[(x- 6)(x+7)=0\]

Answer 2

x=6
x=-7
x=6 is solution

Answer 3

\[\sqrt{9x + 67} = x+ 5\]
squaring both sides
9x + 67 = x² + 10x + 25
x² + 10x - 9x + 25 - 67 = 0
x² + x - 42 = 0

solve n you will get the roots

Answer 4

you have 3 problem?

Answer 5

\[(4\sqrt{m-1}^2)-(\sqrt{15m-1})^2=0^2\]
16(m-1)-15m+1=0
16m-16 -15m +1 =0
m - 15=0
m=15
check:\[\sqrt{15+1}-\sqrt{(15*15) -1}\]\[4-\sqrt{224}\neq 0\]
no solution

Answer 6

@midrul

z = 8? If 6 sqrt(z-8) = 6

Answer 7

z=9

Answer 8

those are my last two i just posted

Answer 9

\[(3+\sqrt{z-8})^2=(z+7)^2\]\[9+6\sqrt{z-8}+z-8=z+7\]\[6\sqrt{z-8}+1=7\]
z is cancle\[6\sqrt{z-8}=7-1\]\[\sqrt{z-8}=\frac{6}{6}\]\[(\sqrt{z-8})^2=(1)^2\]\[z-8=1\]\[z=8+1\]\[z=9\]

Answer 10

check:\[3+\sqrt{9-8}=\sqrt{9+7}\]\[3+1=\sqrt{16}\]\[4=4\]

Answer 11

I was type wrong first line

Answer 12

\[(3+\sqrt{z-8})^2=(\sqrt{x-7})^2\]

Answer 13

I forgot square