Question

How do you solve ${{\left( -3-4x \right)}^{\dfrac{1}{2}}}-{{\left( -2-2x \right)}^{\dfrac{1}{2}}}=1$

Answer 1

Hint: We can solve the given equation by squaring both LHS and RHS , we may have to square both LHS and RHS again to find the value of x. We have to use the property ${{\left( {{a}^{\dfrac{1}{2}}} \right)}^{2}}=a$ . When we write some number power $\dfrac{1}{2}$ then the number must be positive to get a real result.

Complete step by step answer:
The given equation in the question is ${{\left( -3-4x \right)}^{\dfrac{1}{2}}}-{{\left( -2-2x \right)}^{\dfrac{1}{2}}}=1$
We can solve the above equation by squaring both sides , so squaring both sides we get
${{\left( {{\left( -3-4x \right)}^{\dfrac{1}{2}}}-{{\left( -2-2x \right)}^{\dfrac{1}{2}}} \right)}^{2}}=1$
$\Rightarrow \left( -3-4x \right)+\left( -2-2x \right)+2{{\left[ \left( -3-4x \right)\left( -2-2x \right) \right]}^{\dfrac{1}{2}}}=1$
$\Rightarrow -6x-5+2{{\left[ 8{{x}^{2}}+14x+6 \right]}^{\dfrac{1}{2}}}=1$
Adding 6x+5 in both LHS and RHS we get
$\Rightarrow 2{{\left[ 8{{x}^{2}}+14x+6 \right]}^{\dfrac{1}{2}}}=6x+6$
Diving both LHS and RHS by 2 and then squaring both sides we get
$\Rightarrow 8{{x}^{2}}+14x+6=9{{x}^{2}}+18x+9$
Further solving we get
$\Rightarrow {{x}^{2}}+4x+3=0$
Factoring the above quadratic equation we get
$\Rightarrow \left( x+3 \right)\left( x+1 \right)=0$
So the value of x can be -3 or -1
We can check our answers by putting it in the equation and check whether it satisfy or not
So putting x equal to -3 we get
${{\left( -3-4\times -3 \right)}^{\dfrac{1}{2}}}-{{\left( -2-2\times -3 \right)}^{\dfrac{1}{2}}}$
$\Rightarrow {{9}^{\dfrac{1}{2}}}-{{4}^{\dfrac{1}{2}}}=1$
So -3 is correct answer
Now putting x equal to -1 we get
${{\left( -3-4\times -1 \right)}^{\dfrac{1}{2}}}-{{\left( -2-2\times -1 \right)}^{\dfrac{1}{2}}}$
$\Rightarrow {{1}^{\dfrac{1}{2}}}-0=1$
So -1 is also correct answer

Note: In mathematics the square root of any positive real number can not be a negative number. For example the solution of ${{x}^{2}}=9$ can be 3 or -3 but the value of the square root of 9 is not -3 , it is only 3. So write the solution or roots of the equation ${{x}^{2}}=9$ is $\pm \sqrt{9}$ which is 3 and -3.