$${\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}} = {\sqrt{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{19}}}}$$
I am having problems solving this equation, can anybody please be so kind and help me?
Ohh, and the result is $${\mathtt{x}} = {\mathtt{7}}$$.
+130511 These square root equations can be a little "sticky"..... here we go......
√(2x + 4) - √(x + 1) = √( 3x - 19) first, square both sides....this gives
2x + 4 - 2√[(2x + 4)(x + 1)] + x + 1 = 3x - 19 simplify
3x + 5 - 2√[(2x + 4)(x + 1)] = 3x - 19 leave the root on the left and move everything else to the right side by subtrating 3x, 5 from both sides....this gives us
- 2√[(2x + 4)(x + 1)] = -24 divide both sides by -2
√[(2x + 4)(x + 1)] = 12 square both sides
(2x + 4)(x + 1) = 144 simplify
2x^2 + 6x + 4 = 144 subtract 144 from both sides
2x^2 + 6x - 140 = 0 divide through by 2
x^2 + 3x - 70 = 0 factor
(x + 10) (x - 7) = 0 setting each factor to 0, we have posible solutions of x = -10 and x = 7
Notice that -10 makes all three original roots negative and we can't take a square root of a negative number (this is known as an "extraneous" solution)
Check x = 7 in the original problem....
√(2(7) + 4) - √(7 + 1) = √( 3(7) - 19)
√(18) - √(8) = √( 2)
3√(2) - 2√(2) = √( 2)
√( 2) = √( 2)
Yep...that works right out !!!
Be sure to ask if you have questions....these problems are a little difficult.....
+130511 These square root equations can be a little "sticky"..... here we go......
√(2x + 4) - √(x + 1) = √( 3x - 19) first, square both sides....this gives
2x + 4 - 2√[(2x + 4)(x + 1)] + x + 1 = 3x - 19 simplify
3x + 5 - 2√[(2x + 4)(x + 1)] = 3x - 19 leave the root on the left and move everything else to the right side by subtrating 3x, 5 from both sides....this gives us
- 2√[(2x + 4)(x + 1)] = -24 divide both sides by -2
√[(2x + 4)(x + 1)] = 12 square both sides
(2x + 4)(x + 1) = 144 simplify
2x^2 + 6x + 4 = 144 subtract 144 from both sides
2x^2 + 6x - 140 = 0 divide through by 2
x^2 + 3x - 70 = 0 factor
(x + 10) (x - 7) = 0 setting each factor to 0, we have posible solutions of x = -10 and x = 7
Notice that -10 makes all three original roots negative and we can't take a square root of a negative number (this is known as an "extraneous" solution)
Check x = 7 in the original problem....
√(2(7) + 4) - √(7 + 1) = √( 3(7) - 19)
√(18) - √(8) = √( 2)
3√(2) - 2√(2) = √( 2)
√( 2) = √( 2)
Yep...that works right out !!!
Be sure to ask if you have questions....these problems are a little difficult.....
