Question

Use the quadratic formula to solve x2 + 9x + 10 = 0.
What are the solutions to the equation?

Answer 1

`\large\boxed{x=(-9\pm√(41))/(2)}`

Explanation:

`\text{The quadratic formula of}`

`ax^2+bx+c=0`

`\text{If}\ b^2-4a<0,\ \text{then the equation has no solution}\\\\\text{If}\ b^2-4ac=0,\ \text{then the equation has one solution}\ x=(-b)/(2a)\\\\\text{If}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=(-b\pm√(b^2-4ac))/(2a)\\=========================================`

`\text{We have}\ x^2+9x+10=0\\\\a=1,\ b=9,\ c=10\\\\\text{substitute:}\\\\b^2-4ac=9^2-4(1)(10)=81-40=41>0\qquad _{\text{two solutions}}\\\\√(b^2-4ac)=√(41)\\\\x=(-9\pm√(41))/(2(1))=(-9\pm√(41))/(2)`

Answer 2

The solution of given quadratic equation = [-9 ± √41]/2

Explanation:

Points to remember

solution of a quadratic equation ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/2a

It is given that, x² + 9x + 10 = 0.

To find the solution

Here a = 1, b= 9 and c = 10

x = [-b ± √(b² - 4ac)]/2a

= [-9 ± √(9² - 4*1*10)]/2*1

= [-9 ± √(81 - 40)]/2

= [-9 ± √(41)]/2

= [-9 ± √41]/2

Therefore the solution of given quadratic equation = [-9 ± √41]/2