Question

& The roots of the equation x² + 7x +10=0 are
a 2 & 5. b. -2and 5. c. -2 and -5 d. 2and -5​

Answer 1

Answer :-

Roots of the equation x² + 7x + 10 = 0 are - 2 & - 5 : Option c

Explanation :-

Given Equation :

x² + 7x + 10 = 0

On comparing the above equation with ax² + bx + c = 0, We get

• a = 1

• b = 7

• c = 10

Discriminant (D) = b² - 4ac

= (7)² - 4(1)(10)

= 49 - 40

= 9

Here D > 0

So the equation have two distinct real roots

Finding the roots of the equation using quadratic formula

x = (-b ± √D)/2a

By substituting the values

⇒ x = (-7 ± √9)/2(1)

⇒ x = (-7 ± 3)/2

⇒ x = (-7 + 3)/2 or (-7 - 3)/2

⇒ x = - 4/2 or - 10/2

⇒ x = - 2 or - 5

∴ the roots of the equation x² + 7x + 10 = 0 are - 2 & - 5 : Option c

Answer 2

Answer:-

x = -2 and -5

Explanation:-

Given:-

quadratic equation x²+7x+10 = 0

To Find:-

Roots of the equation

Solution:-

Middle-term splitting:-

x²+7x+10 = 0

x² + 5x + 2x +10 = 0

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

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

•°• x = -2 and x = -5

Quadratic Formula:-

$\boxed{x = \dfrac{-b \pm \sqrt{{b}^{2}-4ac}}{2a}}$

Here,

• a = 1
• b = 7
• c = 10

On putting the values

$x = \dfrac{-7 \pm \sqrt{{7}^{2}-4(1)(10)}}{2(1)}$

$x = \dfrac{-7 \pm \sqrt{49-40}}{2(1)}$

$x = \dfrac{-7 \pm \sqrt{9}}{2(1)}$

$x = \dfrac{-7 \pm 3}{2(1)}$

Therefore,

x = -2 and -5 ______option(c)