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Inessa [10]
2 years ago
8

6. Solve by factoring: 2x²-11x+14=0

Mathematics
2 answers:
Papessa [141]2 years ago
7 0

Answer:

(2x-7) (x-2)

Step-by-step explanation:

1. add the numbers

2x²-11x+14+0

2x²-11x+14

Mariana [72]2 years ago
3 0

Step-by-step explanation:

Solve 2x^2-11x+14=0 give me 2 ways to get the 2 values of x?

The first way is to make factors by middle-term splitting. You get (x-2)(2x-7)=0, so the solution is x=2,7/2.

The other way is to use the formula for the solution of roots. Roots= (-b+ rootD)/2a, (-b-rootD)/2a, if the equation is of the form ax^2+bx+c=0. Here D is the discriminant. D=b^2-4ac

How do I solve 12x^2+11x+2=0?

Solve for x. 5x3−2x2−47x−14=0?

How can I prove that x^2+2x+2=0?

Is the sequence X^2 + 2x -2 =0?

What are the steps to solve 2^(2x)-3(2^x) +2=0?

Lets assume your equation is of the form ax^2+ bx + c =0

First method :

By breaking 'b' factor of x into two parts using factors of 'ac' ( a *c)

like 2* 14 = 28 = 7* 4 ( b is 11 which could be splitted in values 7 and 4)

equation becomes 2x^2 -4x-7x+14 = 0

take out common factors 2x(x-2) -7 (x-2) = 0

which is (2x-7) (x-2) =0 ; x =2,7/2 are two values

Second method:

make equation in the form of (x+h)^ 2 - k = 0 ; then x+h = +sqrt(k) and -sqrt(k)

which will give x as -h+sqrt(k) , +sqrt(k)

2x^2 - 11x +14 =0 wil become x^2 -11/2 x + 7 = 0

which is (x-11/4)^2 + 7- 121/16 =0

(x-11/4)^2 = (121/16) - 7 = 9/16

x- 11/4 = 3/4 and -3/4

x = 14/4 and 8/4 = 7/2 and 2

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Jordan has 5/6 dvds as bob, so if jordan has 55 dvds how much does bob have?
Luba_88 [7]

Answer:

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Step-by-step explanation:

Let's assume

Bob has 'x' number of dvds

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3 0
3 years ago
One side of a triangle is 3 times the second side. The third side is 13 feet longer than the second side. The perimeter of a tri
lisabon 2012 [21]

Answer:

  33, 11, 24 feet

Step-by-step explanation:

Let s represent the length of the second side. Then the length of the first side is 3s and the length of the third side is s+13. The perimeter is the sum of side lengths:

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The side lengths are ...

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