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This isn't a quadratic equation, it would need one more x term (2x^2 - 7x + 23). your answer will be : 

2x^2-7 + 23 = 0 

2x^2 +16 = 0 

2x^2=-16 

x = 2√2i and -2√2i

This isn't a quadratic equation, it would need one more x term (2x^2 - 7x + 23). But here is your answer: 

2x^2-7 + 23 = 0 

2x^2 +16 = 0 

2x^2=-16 

x = sqrt8

2x^2=7-23

2x^2= -16

x^2= -16/2

x^2= -8

X= -√8

X= -2√2

Where value of √2= 1.414

X=  +2.828 and -2.828

x=sqaure root of 8

If it is the same equation you have written , then

2x^2-7+23 = 0

=2x^2 +16=0

= 2x^2 = -16

=x^2 = -8

=x=+2.828i,-2.828i

I think it will be 2x^2-7x+23=0, since the highest degree of a variable (here x) is 2 its a quadratic equation, solving this means we must find out those values of x which will satisfies the equation. We can solve this by using quadratic formula.

x=+2.828i,-2.828i

I think your are forgetting to multiply x with 7

Hi Abdul! This is how I would answer this question:

We can simplify this equation into 2x^2     +16      =0 

A quadratic equation looks like this: ax^2   +    bx     + c    =0

The number that goes before the    x^2    is 2, so a=2

There isnt any x's that are not squared, which would be for example 4x, so that means b has to be 0

C is a number that is on its own with no x's, so c must be 16!

Hope this helps 

2x^2-7+23=0 is a quadratic equation. 

Reason:

if compared to the standard form ax^2+bx+c=0 ; a is not equal to zero,

we can find it holds true to be a quadratic eqn with a=2; b=0; c=16

Yes it's a Quadratic equation.

we know that

image: http://www.yourdictionary.com/image/articles/18925.ThinkstockPhotos-502612028.jpg

quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants

Here the power of x is 2 

2x^2-7+23=0. ok so the trick to this one is visualizing the simplified form:

combine like terms 2x^2+16=0

It certainly is a quadratic equation, as Vik mentions the degree of the polynomial is 2. IE the highest order of x is a=2 / 2x^2.

We have been mentioning whether in fact the equation is a=2 b = -7 and c = -16

or a = 2 b = 0 and c = '7-23', in other words, ax^2+bx+c=0.

2x^2 -7+23=0 , yes this is Quadratic Equation because the highest power of (Variable) x is 2.

Ramendu may i say working very clear and accurate, with enough appropriate statement for extensive answer. Example stuff. 

M Cover

Very clearly there is at least a complex solution should as suggested earlier the equation have a term in x not in fact -7 + 23

2x^2=-16   => x^2=-8   this equation has no real solutions!

Pure imaginary roots  x=+/-2sqrt2i

I may finally mention, as I should consider that the question is perhaps a typo in form of ax2 - bx + c = 0, where complex analysis has replaced correctly the prev. mention (exactly a=2 b=-7 and c=23)

https://www.symbolab.com/solver/step-by-step/2x%5E%7B2%7D-7x%2B23%3D0 

The link further directs into complex mathematics and explains thoroughly should the equation have read 2x^2-7x+23=0 (which is more difficult as exists another substituted variable x)


The answer is listed in Complex form, however between lines 2 and 3 of the last answer 2x^2 onto x^2 in relation to similar operation -16, should map onto -8 as Maria S mentions primarily.

It is an easy mistake to make when trying to break into negative quantity to arrange into the same function as the approximate amount for i. Particularly, the power operation present and the x co-efficiant has swung in both being  the number 2. Working clearly from line 1 introducing a negative function 

RHS to reach an suitable arrangement into i is also clear and commendable, certainly the only suitable way to answer totally correctly. @x^2=-8, with complex procedure you may deduce approximately the roots 2.2^1/2i & -2.2^1/2i for x

2x^2 = 7-23

2x^2 = -16

x^2 = -4

x = 2i

Check the question again please, you might have a wrong sign somewhere. If you solve the above equation you get 2x^2=-23+7, hence 2x^2=-16, so x^2=-8, which is an equation that doesn't have a solution within the set of the real numbers. However, if you have learned complex analysis it can be solved and I can send you the answer. The solution is an imaginary number.

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