Question

How do you solve
3x^2 + 29x -10=0

Answer 1

`\sf\\\textsf{Here, you need to find two numbers whose difference is 29 and whose product is}\\\textsf{3(10)=30. They are 30 and 1 because 30 - 1 = 29 and 30}*1=30.\\\textsf{Now we write 29 as }30-1.\\3x^2+29x-10=0\\or,\ 3x^2+(30-1)x-10=0\\or,\ 3x^2+30x-x-10=0\\or,\ 3x(x+10)-1(x+10)=0\\or,\ (x+10)(3x-1)=0\\i.e.\ x=-10\ or\ 1/3`

General rule for solving quadratic equation ax² + bx + c = 0:

i. If constant (c) is positive:

Find two numbers p and q such that p + q = b and pq = ac.

ii. If constant (c) is negative:

Find two numbers p and q such that p - q = b and pq = -ac.

In the above question, constant term i.e. -10 was negative number. So, we searched for two numbers whose difference is 29 and whose product is 30 as per the rule.

Answer 2

The solutions for the given equation are x = -10/3 and x = 1. To solve the equation 3x^2 + 29x -10 = 0, use the quadratic formula.

Step-by-step explanation:

To solve the equation 3x^2 + 29x -10 = 0, we can use the quadratic formula. The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the given equation, a = 3, b = 29, and c = -10. Substituting these values into the quadratic formula:

x = (-29 ± √(29^2 - 4 * 3 * -10)) / (2 * 3)

Simplifying further:

x = (-29 ± √(841 + 120)) / 6

x = (-29 ± √961) / 6

x = (-29 ± 31) / 6

Therefore, the solutions for the given equation are x = -10/3 and x = 1.