Question

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Answer 1

Question :

• Factorise : x² + 6x + 10 = 0

Solution :

• x² + 6x + 10 = 0

In the above equation

• a ( coefficient of x²) = 1
• b ( coefficient of x) = 6
• c ( constant term) = 10

Now , find the discriminant of the equation :

• Discriminant = b² - 4ac

= 6² - 4 × 1 × 10

= 36 - 40

= -4

Now , using the quadratic formula to find the root of the equation .

• The quadratic formula is given by :

$\boxed{ \implies\tt x = \frac{ - b \pm \sqrt{discriminant } }{2a}}$

$\tt x = \dfrac{ - 6 + \sqrt{ - 4} }{2 \times 1} \qquad \: x = \dfrac{ - 6 - \sqrt{ - 4} }{2 \times 1}$

$\tt x = \dfrac{ - 6 + 2i}{2} \qquad x = \dfrac{ - 6 - 2i}{2}$

• where i = - 1

$\tt x = \dfrac{ -6 + 2 \times -1}{2} \qquad x = \dfrac{ - 6 - 2 \times -1}{2}$

$\tt x = \dfrac{-6 - 2}{2} \qquad x = \dfrac{-6 + 2}{2}$

$\tt x = \dfrac{-8}{2} \qquad x = \dfrac{ -4}{2}$

$\tt x = -4 \qquad x = -2$

Answer 2

★FIND:

The value of 'x'

★GIVEN,

$\bold{x^2+6x+10=0}$

★SOLUTION:

$\bold{x^2+6x+10=0}$

$\bold{where, }$

$\bold{a=1,}$ $\bold{b=6,}$ $\bold{c=10}$

$x = \frac{ - b + - \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - 6 + - \sqrt{36 - 4(1)(10)} }{2(1)} \\ = \frac{ - 6 + - \sqrt{36 - 40} }{2} \\ = \frac{ - 6 + - \sqrt{ - 4} }{2} \\ = \frac{ - 6 + - \sqrt{4i {}^{2} } }{2} \\ (i {}^{2} = - 1) \\ = \frac{ - 6 + - \sqrt{4}i }{2} \\ = \frac{ - 6 + - 2i}{2} \\ = - 3 + - i \\ x = - 3 + - i$