Question

The roots of the quadratic equation x2+5x-(alpha +1)(alpha+6)=0 where alpha is a constant

Answer 1

Answer: (alpha + 1 ), - ( alpha + 6 )

Step-by-step explanation:

Answer 2

Find the roots of the quadratic equation.

Explanation:

• The quadratic equations of the form where a, b and c are real numbers has roots $x_1,\ x_2$ that are given by,                                                 $x_1=\frac{-b+\sqrt{b^2-4ac} }{2a} ,\ x_2=\frac{-b-\sqrt{b^2-4ac} }{2a}$  
• Given is the quadratic equation $x^2+5x-(\alpha +1)(\alpha +6)=0$ here alpha is constant then we have, $a=1,\ b=5,\ c=-(\alpha+1)(\alpha+6)$ .
• Hence from the standard quadratic equation mentioned above we get the roots of the equation as,  $x_1=\frac{-5+\sqrt{5^2+4(\alpha+1)(\alpha+6)} }{2} ,\ x_2=\frac{-5-\sqrt{5^2+4(\alpha+1)(\alpha+6)} }{2}\\->x_1=\frac{-5+\sqrt{25+4\alpha^2+28\alpha+24} }{2} ,\ x_2=\frac{-5-\sqrt{25+4\alpha^2+28\alpha+24} }{2}\\->x_1=\frac{-5+\sqrt{4\alpha^2+28\alpha+49} }{2},\ x_2=\frac{-5+\sqrt{4\alpha^2+28\alpha+49} }{2}\\->x_1=\frac{-5+\sqrt{(2\alpha+7)^2} }{2},\ x_2=\frac{-5-\sqrt{(2\alpha+7)^2} }{2} \\->x_1=\frac{-5+2\alpha+7}{2} ,\ x_2=\frac{-5-2\alpha-7}{2} \\->x_1=\alpha+1,\ x_2=-(\alpha+6)$          
• Hence the roots of the given equation are $-(\alpha+6)\ and\ (\alpha+1)$.