Question

Solve. 10x^2 = 6 + 9x10x 2 =6+9x10, x, start superscript, 2, end superscript, equals, 6, plus, 9, x Choose 1 answer: Choose 1 answer: (Choice A) A x =\dfrac{5 \pm \sqrt{65}}{-2}x= ?2 5± 65 ? ? x, equals, start fraction, 5, plus minus, square root of, 65, end square root, divided by, minus, 2, end fraction (Choice B) B x =\dfrac{9 \pm \sqrt{321}}{20}x= 20 9± 321 ? ? x, equals, start fraction, 9, plus minus, square root of, 321, end square root, divided by, 20, end fraction (Choice C) C x =\dfrac{4 \pm \sqrt{26}}{10}x= 10 4± 26 ? ? x, equals, start fraction, 4, plus minus, square root of, 26, end square root, divided by, 10, end fraction (Choice D) D x =\dfrac{-1 \pm \sqrt{109}}{18}x= 18 ?1± 109 ? ?

Answer 1

B

Explanation:

I got it right

Answer 2

Option B.

Explanation:

If a quadratic equation is defined as

`ax^2+bx+c=0` .... (1)

then the quadratic formula is

`x=(-b\pm √(b^2-4ac))/(2a)`

The given equation is

`10x^2=6+9x`

It can rewritten as

`10x^2-9x-6=0` .... (2)

On comparing (1) and (2) we get

`a=10,b=-9,c=-6`

Using quadratic formula we get

`x=(-(-9)\pm √((-9)^2-4(10)(-6)))/(2(10))`

`x=(9\pm √(81+240))/(20)`

`x=(9\pm √(321))/(20)`

Therefore, the correct option is B.