solutions of quadratic equations

**daniel123** · September 26th 2010, 11:41 AM

Hi, i am struggling with factorising the quadratic equations. I have got to grips with basic factorisation. any help?

1. solve the following quadratic equations by factorisation:

a)
(x+6)(x-3)=22
......
i have tried this one and came out with this answer:
x(x-3)+6(x-3)=22
x^2-3x+6x-18=22
x^2+3x-22-16=0
x^2+3x-38
(x+19)(x-2)

Is this correct?

b) 20x=4+3/x

c) 4t(3t-2)=3-8t

2.
a) x^2=7x+4

b) (3x+4)(x-1)=1

c) 1/2x-7=x/x-1

**skeeter** · September 26th 2010, 11:53 AM

Originally Posted by daniel123

Hi, i am struggling with factorising the quadratic equations. I have got to grips with basic factorisation. any help?

1. solve the following quadratic equations by factorisation:

a)
(x+6)(x-3)=22

x^2 + 3x - 18 = 22

x^2 + 3x - 40 = 0

(x + 8)(x - 5) = 0

set each factor = 0 and solve.

b) 20x=4+3/x

multiply every term by x ...

20x^2 = 4x + 3

20x^2 - 4x - 3 = 0

factor, set each factor = 0 and solve

c) 4t(3t-2)=3-8t

12t^2 - 8t = 3 - 8t

set = 0 and solve as before

2.
a) x^2=7x+4

set = 0 and solve as before

b) (3x+4)(x-1)=1

expand, set = 0 and solve

c) 1/(2x-7)=x/(x-1)

note the parentheses.

cross multiply ...

x(2x-7) = 1(x-1)

distribute, set = 0 and solve

...

**daniel123** · September 26th 2010, 01:31 PM

i have the answers to 1.
just to check im going in the right direction
a)factors x=-8 x=5
b)factors x=3/4 x=-1/5
c)factors x=-3/3 x=1/4

I am struggling still with question 2.
so far i have worked these out as fractions. i need to change them into an answer with 3 decimal places, but i do not know if i have the correct answer....
a)factors x=4 x=-1
b)factors x=5/3 x=-1
c)factors x=-1/2 x=-1

**pickslides** · September 26th 2010, 02:12 PM

Originally Posted by daniel123

2.
a) x^2=7x+4

$\displaystyle x^2=7x+4$

$\displaystyle x^2-7x-4=0$

Completing the square

$\displaystyle x^2-7x+\frac{49}{4}-4-\frac{49}{4}=0$

$\displaystyle \left(x-\frac{7}{2}\right)^2-4-\frac{49}{4}=0$

$\displaystyle \left(x-\frac{7}{2}\right)^2-\frac{65}{4}=0$

$\displaystyle \left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{65}}{2}\right)^2=0$

Now apply the difference of 2 squares

Originally Posted by daniel123

b) (3x+4)(x-1)=1

Expand this one out first and make RHS = 0, then have another attempt.

Originally Posted by daniel123

c) 1/2x-7=x/x-1

What is this equation?

$\displaystyle \frac{1}{2x-7} = \frac{x}{x-1}$

or

$\displaystyle \frac{1}{2x}-7 = \frac{x}{x}-1$

Use brackets to provide clarity

**daniel123** · September 27th 2010, 08:37 AM

I have tried an alternate way shown in an exercise book for question 2. I have attemted a) . Just woundering if i was going in the right direction?
2.
a)x^2=7x+4
b)(3x+4)(x-1)=1
c)1/(2x-7) = x/(x-1)

2.
a)
Roots are x=7.531 x=-0.531

**pickslides** · September 27th 2010, 02:03 PM

Originally Posted by daniel123

2.
a)
Roots are x=7.531 x=-0.531

These look good.

Originally Posted by daniel123

b)(3x+4)(x-1)=1

Did you expand this one yet?

Originally Posted by daniel123

c)1/(2x-7) = x/(x-1)

$\displaystyle \frac{1}{2x-7} = \frac{x}{x-1}$

Cross multiplying

$\displaystyle x-1 = x(2x-7)$

$\displaystyle x-1 = 2x^2-7x$

$\displaystyle 2x^2-8x+1=0$

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