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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