Question 19

The number of real solutions of the equation $$x^2 - 10 \mid x \mid - 56 = 0$$ is

The number of real solutions of the equation $$x^2 - 10 \mid x \mid - 56 = 0$$ is

We need to find the solutions to the equation - $$x^2 - 10 \mid x \mid - 56 = 0$$

Taking two cases

Case 1: x > 0

$$x^2-10x-56=0$$

$$x^2-14x+4x-56=0$$

$$x\left(x-14\right)+4\left(x-14\right)=0$$

$$\left(x+4\right)\left(x-14\right)=0$$

x = -4 and x = 14

Since, x > 0

x = 14 is the only possible value.

Case 2: x < 0

$$x^2+10x-56=0$$

$$x^2+14x-4x-56=0$$

=> $$x\left(x+14\right)-4\left(x+14\right)=0$$

=> $$\left(x-4\right)\left(x+14\right)=0$$

x = 4 and x = -14

Since, x < 0

x = -14 is the only possible value.

Therefore, there are 2 possible real solutions.

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