Question

If a and are the roots of quadratic equation such that, a+b=1 and a-b=5 find the quadratic equation.

Answer 1

COMPLETE QUESTION-

• If a and b are the roots of quadratic equation such that, a+b=1 and a-b=5. Find the quadratic equation.

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Quadratic\:equation=x^{2}-x-6}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{Given : } \\ \implies a \: \: and \: \: b \: \in(a {x}^{2} + bx + c = 0) \\ \\ \implies a + b =1 \\ \\ \implies a - b = 5 \\ \\ \underline \bold{To \: Find : } \\ \implies Quadratic \: equation = ?$

• According to given question :

$\bold{Using \: substitution \: method : } \\ \implies a + b = 1 \\ \\ \implies a = 1 - b - - - - - (1) \\ \\ \bold{Substituting\: value \: of \:a \: in \: second \: equation} \\ \implies a - b = 5 \\ \\ \implies 1 - b - b = 5 \\ \\ \implies - 2b = 4 \\ \\ \implies b \frac{ - 4}{2} \\ \\ \bold{\implies b = - 2} \\ \\ \bold{Putting \:value \: of \: b \: in \: (1)} \\ \implies a = 1 - b \\ \\ \implies a = 1 - ( - 2) \\ \\ \bold{\implies a =3 } \\ \\ \bold{For \: Quadratic \: equation} \\ \implies {x}^{2} - (a + b)x + ab \\ \\ \implies {x}^{2} - 1x + 3 \times ( - 2) \\ \\ \bold{\implies {x}^{2} - x - 6 }$

Answer 2

$\bold{using \: substitution \: method : } \\ \to a + b = 1 \\ \\ \to a = 1 - b - - - - - (1) \\ \\ \bold{substituting\: value \: of \:a \: in \: second \: equation} \\ \to a - b = 5 \\ \\ \to 1 - b - b = 5 \\ \\ \to - 2b = 4 \\ \\ \to b \frac{ - 4}{2} \\ \\ \bold{\to b = - 2} \\ \\ \bold{putting \:value \: of \: b \: in \: (1)} \\ \to a = 1 - b \\ \\ \to a = 1 - ( - 2) \\ \\ \bold{\to a =3 } \\ \\ \bold{for \: quadratic \: equation} \\ \to {x}^{2} - (a + b)x + ab \\ \\ \to {x}^{2} - 1x + 3 \times ( - 2) \\ \\ \bold{\to {x}^{2} - x - 6 }$