Question

The value of $ x $ which satisfies $ x\left| x \right| + 7x - 8 = 0 $ is
A. $ x = 1 $
B. $ x = 0,1 $
C. $ x = 1,2 $
D. $ x = - 1 $

Answer 1

Hint: Here we are given an equation $ x\left| x \right| + 7x - 8 = 0 $ and we are asked to find the value $ x $ that satisfies $ x\left| x \right| + 7x - 8 = 0 $ . That is we need to calculate the solution of the given equation.

Formula used:
The formula to be used to find the solutions of the quadratic equation\[a{x^2} + bx + c = 0\;\] is as follows.
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $

Complete step by step solution:
The given equation is $ x\left| x \right| + 7x - 8 = 0 $ . Here we are asked to find the solution of the given equation.
Let us deal it with cases.
Case a:
The given equation is $ x\left| x \right| + 7x - 8 = 0 $ .
Let us assume that $ x $ is greater than zero. That is $ x > 0 $ .
We know that the absolute value $ x $ is equal to $ x $ when $ x $ is greater than zero.
That means $ \left| x \right| = x $ when $ x > 0 $ ….. $ \left( 1 \right) $
Let us substitute $ \left( 1 \right) $ in the given equation.
 $ x \times x + 7x - 8 = 0 $
 $ \Rightarrow {x^2} + 7x - 8 = 0 $
Now, we shall split the middle term of the above equation.
 $ \Rightarrow {x^2} + 8x - x - 8 = 0 $
 $ \Rightarrow \left( {x + 8} \right)x - \left( {x + 8} \right) = 0 $
 $ \Rightarrow \left( {x - 1} \right)\left( {x + 8} \right) = 0 $
 $ \Rightarrow x - 1 = 0 $ or $ x + 8 = 0 $
Thus we get $ x = 1orx = - 8 $
But we have assumed that $ x $ is greater than zero. So $ x = - 8 $ is not possible.
Therefore $ x = 1 $ is the required solution.
Case b:
The given equation is $ x\left| x \right| + 7x - 8 = 0 $ .
Let us assume that $ x $ is less than zero. That is $ x < 0 $ .
We know that the absolute value $ x $ is equal to $ - x $ when $ x $ is less than zero.
That means $ \left| x \right| = x $ when $ x < 0 $ ….. $ \left( 2 \right) $
Let us substitute $ \left( 2 \right) $ in the given equation.
 $ x\left( { - x} \right) + 7x - 8 = 0 $
 $ \Rightarrow - {x^2} + 7x - 8 = 0 $
 $ \Rightarrow {x^2} - 7x + 8 = 0 $ (Here we have multiplied throughout by the sign $ - $ )
Now we shall apply the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ to find the solutions of the above quadratic equation.
 $ x = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4 \times 1 \times 8} }}{{2 \times 1}} $
 $ \Rightarrow x = \dfrac{{7 \pm \sqrt {49 - 32} }}{2} $
 $ \Rightarrow x = \dfrac{{7 \pm \sqrt {17} }}{2} $
 $ \Rightarrow x = \dfrac{{7 + \sqrt {17} }}{2} $ or $ x = \dfrac{{7 - \sqrt {17} }}{2} $
Thus, we get $ x = 5.56 $ or $ x = 1.43 $ .
But we have assumed that $ x $ is less than zero. So $ x = 5.56 $ or $ x = 1.43 $ is not possible.
Therefore there is no solution when $ x $ is less than zero.
Hence we get only one solution for the given equation.
Thus $ x = 1 $ is the solution for $ x\left| x \right| + 7x - 8 = 0 $ and option A is the correct.

Note:
The absolute value of $ x $ is equal to $ x $ when $ x $ is greater than zero, the absolute value of $ x $ is equal to $ - x $ when $ x $ is less than zero, and the absolute value of $ x $ is equal to zero when $ x $ is equal to zero. That means $ \left| x \right| = x $ when $ x > 0 $ , $ \left| x \right| = x $ when $ x < 0 $ , and $ \left| x \right| = 0 $ when $ x = 0 $ .