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Question: The sum of the roots of Q.E, \({x^2} = \left| {x - 6} \right|\) is –k. The value of k is......

The sum of the roots of Q.E, x2=∣x−6∣{x^2} = \left| {x - 6} \right| is –k. The value of k is...

Explanation

Solution

In this particular type of question use the concept that if there is any modulus part in the equation equate the function which is inside the modulus to zero and evaluate the condition in which function opens as positive and in which function opens as negative, then factorize the equation so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given quadratic equation:
x2=∣x−6∣{x^2} = \left| {x - 6} \right|
Now as we see that the R.H.S part of the above function has modulus.
So there is a condition substitute x – 6 = 0
Therefore, x = 6.
So when x > 6 modulus functions open as positive and when x < 6 modulus functions open as negative.
So when, x > 6
Therefore, x2=+(x−6){x^2} = + \left( {x - 6} \right)
⇒x2−x+6=0\Rightarrow {x^2} - x + 6 = 0
Now factorize the equation we have,
⇒x2−3x+2x+6=0\Rightarrow {x^2} - 3x + 2x + 6 = 0
⇒x(x−3)−2(x−3)=0\Rightarrow x\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0
⇒(x−3)(x−2)=0\Rightarrow \left( {x - 3} \right)\left( {x - 2} \right) = 0
⇒x=3,2\Rightarrow x = 3,2
So we get the solution which is less than 6, so x can never be greater than 6.
So the above condition is false.
So when, x < 6
Therefore, x2=−(x−6){x^2} = - \left( {x - 6} \right)
⇒x2+x−6=0\Rightarrow {x^2} + x - 6 = 0
Now factorize the equation we have,
⇒x2+3x−2x−6=0\Rightarrow {x^2} + 3x - 2x - 6 = 0
⇒x(x+3)−2(x+3)=0\Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0
⇒(x+3)(x−2)=0\Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0
⇒x=−3,2\Rightarrow x = - 3,2
So we get both the solutions less than 6.
So the above condition is true.
So the required roots of the quadratic equation is, -3 and 2.
Now the sum of the roots is (-3 + 2) = -1
Now it is given that the sum of the roots is –k, so on comparing, k = 1.
So the value of k = 1.

Note :Whenever we face such types of questions the key concept we have to remember is that the quadratic formula to solve complex quadratic equation i.e. whose factorization is not possible, so compare the coefficients of the quadratic equation with standard equation and substitute then I the formula as above and simplify, if we get un equal but real roots then we can say the roots are real, if we get equal and real then we can say roots are equal, if we get complex roots then we can say that the roots of the quadratic equation are complex (i.e. a combination of real and imaginary).