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Question

Find the number of real solution to the quadratic equation
$$-(x+2)(x-1)=3$$


Solution

Solve the quadratic equation $$-(x+2)(x-1)=3$$ as follows:
$$-(x+2)(x-1)=3$$
$$\Rightarrow -(x^2-x+2x-2)=3$$
$$\Rightarrow -(x^2+x-2)=3$$
$$\Rightarrow -x^2-x-1=0$$
$$\Rightarrow x^2+x+1=0$$
Finding the roots of the above quadratic equation $$x^2+x+1=0$$ as shown below:
$$x=\dfrac { -1\pm \sqrt { 1-4 }  }{ 2 } =\dfrac { -1\pm \sqrt { -3 }  }{ 2 } =\dfrac { -1\pm \sqrt { 3i }  }{ 2 }$$
The quadratic equation has imaginary roots and hence, it has no real roots. 

Maths

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