Solve. x^2 - 5x - 3 = 0?

Question

Solve.
1. x^2 - 5x - 3 = 0

Solve.

2. x^2 - 10x + 9 =0

Answer 1

1) x² - 5x - 3 = 0
x = 5 /2 ± (√25 - 12) / 2
x = 5.54 or -0.54

2) x² - 10x+ 9= 0
x = 10/2 ±(√(100 - 36))/2
x = 5 ± 4
x = 9 or 1

Answer 2

#2 can be easily factored, so I'll get that one out of the way first:
(x - 9)(x - 1) = 0
x = 1, 9

As for #1, lots of people have suggested using the quadratic equation, which is wonderful _as long as you remember it_. For my purposes, I much prefer a method called "completing the square." It's logical enough to where, even if you forget the steps, you can work through them. It also teaches you how to solve problems rather than just plugging in numbers. Anyway...

This is a good equation to work out by completing the square. To do this, take the following steps:

1. Move everything but the constant to one side of the equation:
x^2 - 5x = 3

2. Take half of the coefficient of the x-term, and square it. Add this square to both sides of the equation:
x^2 -5x + 6.25 = 3 + 6.25 (-2.5^2 = 6.25)

3. Convert the left side to a square (x + the number you just squared):
(x - 2.5)^2 = 9.25

4. Square-root both sides, remembering the "±" on the right-hand side. Simplify as necessary:
x - 2.5 = ±(sqrt)9.25

5. Solve for x
x = 2.5 ±(sqrt)9.25

Remember that the "±" means that you have two answers, not one.

Answer 3

(1) x² - 5x - 3 = 0

For this one, I don't see any factoring (I could easily be wrong), so I will use the quadratic formula.

Quadratic Formula (for an equation ax² + bx + c = 0)
x = [-b ± √(b² - 4ac)] / 2a
x = [-(-5) ± √((-5)² - 4(1)(-3))] / 2(1)
x = [5 ± √(25 - (-12))] / 2
x = [5 ± √(37)] / 2
x = [5 ± (6.08)] / 2
x = (5 + 6.08) / 2 = 11.08 / 2 = 5.54
x = (5 - 6.08) / 2 = -1.08 / 2 = -0.54
x = 5.54 or x = -0.54

(2) x² - 10x + 9 = 0
This one is factorable since (-9)(-1) = 9 and (-9) + (-1) = 10...
(x - 9)(x - 1) = 0
x - 9 = 0
x = 9
x - 1 = 0
x = 1

x = 9 or x = 1

Answer 4

We need to use the quadratic formula to solve this since we can't factor the equation easily. The quadratic formula is: -b +/- srt(b^2 - 4ac) / 2a in this case a = 1, b = -5 and c = -3 5 +/- sqrt(-5^2 - 4*1*(-3)) / 2*1 = 5 +/- sqrt (25 - (-12)) /2 = 5 +/- sqrt (37) /2 so the two roots of the equation are: 5 + sqrt(37) /2 = 5.54138 5 - sqrt(37) / 2 = -0.54138

Answer 5

First one needs the quadratic formula
x = (-b +or- sqrt(b^2 - 4ac))/(2a), with a = 1, b = -5, c = -3.

Second one is easily done by factors: Think of two numbers which multiply together to give +9 and add together to give -10 [Hint: they're both negative], then find the two values of x which, in turn, make the factors zero.

Answer 6

In both cases use the quadratic formula. i just did one for you and failed to write the formula. Here is is.

x= [-b +/- sqrt b^2 -4ac)]/2a

a=the coefficient of x^2, b= the coefficient of x with the sign, and c is the constant, with the sign.


In your first example a=1,b=-5, and c=-3

In your second example, a=1, b=-10, and c=+9


ACTUALLY, THE SECOND ONE CAN BE FACTORED. iT'S A LOT EASIER THAT WAY.
good luck

Answer 7

By using Quadratic formula

a = 1, b = -5, c = -3.

x = (-b +or- sqrt(b^2 - 4ac))(2a)

the answers are X1 =4.302, X2 = 0.6975

Answer 8

1.)
x^2 - 5x - 3 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-5) ± sqrt((-5)^2 - 4(1)(-3)))/(2(1))
x = (5 ± sqrt(25 + 12))/2
x = (5 ± sqrt(37))/2

x = (1/2)(5 ± sqrt(37))

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2.)
x^2 - 10x + 9 = 0
(x - 9)(x - 1) = 0
x = 9 or 1

Answer 9

using the quadratic formula the answer is (-5 + square root of 37)/2 and (-5 - square root of 37)/2

the second one is (x-1) (x-9)

Answer 10

1. 32
2. 47