Question

Solve the equation x^2

-3x-4=0 by using the formula​

Answer 1

GIVEN :

• $\bf {x^2-3x-4=0}$

TO FIND :

• The roots.

SOLUTION :

Let the roots are α and β.

Here ,

• a = 1

• b = -3

• c = -4

Finding α :

$\longrightarrow\bf \alpha={\dfrac {-b+\sqrt {b^2-4ac}}{2a}}$

$\longrightarrow\sf \alpha = \dfrac { - ( - 3)+\sqrt{(-3)^2 - 4 \times 1 \times( - 4) }}{2 \times 1}$

$\longrightarrow\sf \alpha = \dfrac {3 +\sqrt {9 + 16}}{2}$

$\longrightarrow\sf \alpha = \dfrac {3+\sqrt {25}}{2}$

$\longrightarrow\sf \alpha = \dfrac {3+5}{2}$

$\longrightarrow\sf \alpha = \cancel{\dfrac {8}{2}}$

$\longrightarrow \underline{ \huge{ \boxed{\boxed{ \bf{ \green{\alpha = 4}}}}}}$

Finding β :

$\longrightarrow\bf \beta={\dfrac {-b-\sqrt {b^2-4ac}}{2a}}$

$\longrightarrow\sf \beta = \dfrac { - ( - 3)-\sqrt{(-3)^2 - 4 \times 1 \times( - 4) }}{2 \times 1}$

$\longrightarrow\sf \beta = \dfrac {3 -\sqrt {9 + 16}}{2}$

$\longrightarrow\sf \beta = \dfrac {3-\sqrt {25}}{2}$

$\longrightarrow\sf \beta = \dfrac {3-5}{2}$

$\longrightarrow\sf \beta= \cancel{\dfrac {-2}{2}}$

$\longrightarrow \underline{ \huge{ \boxed{\boxed{ \bf{ \green{\beta = -1}}}}}}$