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

How do you solve (−3−4x)12−(−2−2x)12=1{{\left( -3-4x \right)}^{\dfrac{1}{2}}}-{{\left( -2-2x \right)}^{\dfrac{1}{2}}}=1

Explanation

Solution

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 (a12)2=a{{\left( {{a}^{\dfrac{1}{2}}} \right)}^{2}}=a . When we write some number power 12\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 (−3−4x)12−(−2−2x)12=1{{\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
((−3−4x)12−(−2−2x)12)2=1{{\left( {{\left( -3-4x \right)}^{\dfrac{1}{2}}}-{{\left( -2-2x \right)}^{\dfrac{1}{2}}} \right)}^{2}}=1
⇒(−3−4x)+(−2−2x)+2[(−3−4x)(−2−2x)]12=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
⇒−6x−5+2[8x2+14x+6]12=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
⇒2[8x2+14x+6]12=6x+6\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
⇒8x2+14x+6=9x2+18x+9\Rightarrow 8{{x}^{2}}+14x+6=9{{x}^{2}}+18x+9
Further solving we get
⇒x2+4x+3=0\Rightarrow {{x}^{2}}+4x+3=0
Factoring the above quadratic equation we get
⇒(x+3)(x+1)=0\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
(−3−4×−3)12−(−2−2×−3)12{{\left( -3-4\times -3 \right)}^{\dfrac{1}{2}}}-{{\left( -2-2\times -3 \right)}^{\dfrac{1}{2}}}
⇒912−412=1\Rightarrow {{9}^{\dfrac{1}{2}}}-{{4}^{\dfrac{1}{2}}}=1
So -3 is correct answer
Now putting x equal to -1 we get
(−3−4×−1)12−(−2−2×−1)12{{\left( -3-4\times -1 \right)}^{\dfrac{1}{2}}}-{{\left( -2-2\times -1 \right)}^{\dfrac{1}{2}}}
⇒112−0=1\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 x2=9{{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 x2=9{{x}^{2}}=9 is ±9\pm \sqrt{9} which is 3 and -3.