# -sqrt2, 6sqrt2 - write a quadratic equation in variable x having given numbers as solutions?

x= -sqrt2 or x= 6sqrt2

x+sqrt2 = 0 or

x-6sqrt2 = 0

(x+sqrt2)(x-6sqrt2)

this is where I get stuck - the equation is to be written in standard form ax^2+bx+c = 0

### 5 Answers

- Anonymous1 decade agoBest Answer
Actually you have done what most people find the 'difficult bit!

Now multiply out the brackets:

(x + sqrt 2)(x - 6sqrt 2) = 0

x(x - 6sqrt 2) + sqrt 2 (x - 6sqrt 2) = 0

x^2 - x*6*sqrt2 + x*sqrt 2 - sqrt2*6*sqrt 2 = 0

x^2 - 6sqrt2*x + sqrt 2*x - 6*2 = 0

x^2 - 5sqrt2*x - 12 = 0

x^2 - 5sqrt2*x - 12 = 0

x^2 - 5sqrt2*x - 12

which is now in the form ax^2 + bx + c = 0 where

a = 1, b = -5sqrt2 and c = 12

b has to be left in surd form as shown.

- menziesLv 43 years ago
through fact the answer are -5 and 3. Then enable x=-5 and 3. it is x=-5 and x=3. that's written has x+5=0 and x-3=0. when you consider that the two are equated to 0, consequently (x+5)(x-3)=0. Open the bracket: x^2-3x+5x-15=0. consequently the quadratic equation=x^2+2x-15=0.

- ericlordLv 41 decade ago
(x + √2)(x - 6√2) = x² - (5√2)x - 12 = 0

This IS the "standard form" ax² + bx+ c = 0.

b ISN'T AN INTEGER!

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