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

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 $x = 0$ .