Question

0X square + 3 x minus 10 equal to zero ​

Answer 1

GIVEN :

• $\bf {x^2+3x-10=0}$

TO FIND :

• The roots.

SOLUTION :

Let the roots are α and β.

Here ,

• a = 1

• b = 3

• c = -10

Finding α :

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

$\longrightarrow\sf \alpha = \dfrac { - 3+\sqrt{(3)^2 - 4 \times 1 \times (-10) }}{2 \times 1}$

$\longrightarrow\sf \alpha = \dfrac {-3 +\sqrt {9 + 40}}{2}$

$\longrightarrow\sf \alpha = \dfrac {-3+\sqrt {49}}{2}$

$\longrightarrow\sf \alpha = \dfrac {-3+7}{2}$

$\longrightarrow\sf \alpha = \cancel{\dfrac {4}{2}}$

$\longrightarrow \underline{ \huge{ \boxed{\boxed{ \bf{ \green{\alpha = 2}}}}}}$

Finding β :

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

$\longrightarrow\sf \beta = \dfrac { - 3 -\sqrt{(3)^2 - 4 \times 1 \times( - 10) }}{2 \times 1}$

$\longrightarrow\sf \beta = \dfrac {-3 -\sqrt {9 + 40}}{2}$

$\longrightarrow\sf \beta = \dfrac {-3-\sqrt {49}}{2}$

$\longrightarrow\sf \beta = \dfrac {-3-7}{2}$

$\longrightarrow\sf \beta= \cancel{\dfrac {-10}{2}}$

$\longrightarrow \underline{ \huge{ \boxed{\boxed{ \bf{ \green{\beta = -5}}}}}}$

Answer 2

Step-by-step explanation:

${x}^{2} + 3x - 10 = 0$

${x}^{2} + (5 - 2)x - 10 = 0$

${x}^{2} + 5x - 2x - 10 = 0$

$x(x + 5) - 2(x + 5) = 0$

$(x + 5)(x - 2) = 0$

either, x+5=0, x= -5

or, x-2=0, x = 2