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the sum of roots of an equation is 1 and their product is-6. Write the equation.​

Answer 1

Appropriate Question :-

If the sum of roots of quadratic equation is 1 and their product is-6. Write the quadratic equation.

$\large\underline{\sf{Solution-}}$

Given that,

Sum of the roots of the equation is 1

and

Product of the roots of the equation is - 6.

Let we assume that the

$\rm :\longmapsto\: \alpha \: and \: \beta \: be \: the \: roots \: of \: quadratic \: equation.$

So, we have

$\rm :\longmapsto\: \alpha + \beta = 1$

and

$\rm :\longmapsto\: \alpha \beta = - \: 6$

We know,

The quadratic equation is given by

$\rm :\longmapsto\: {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

On substituting the values, we get

$\rm :\longmapsto\: {x}^{2} - (1)x - 6 = 0$

$\bf\implies \: {x}^{2} - x - 6 = 0$

Verification :-

The quadratic equation is,

$\bf\implies \: {x}^{2} - x - 6 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \: \dfrac{( - 1)}{1} = 1$

and

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha\beta = \: \dfrac{( - 6)}{1} = - \: 6$

Hence, Verified