# Proving there are no roots to a quadratic equation

• September 3rd 2011, 04:55 AM
mabentley
Proving there are no roots to a quadratic equation
Hello everyone,

I understand perfectly how to solve quadratic equations, but have come across a question which somewhat confuses me. I hope someone can help:

"Show that any root of the equation 5 + x - sqrt(3 + 4x) = 0 is also a root of the equation x^2 + 6x + 22 = 0. Hence show that the equation 5 + x - sqrt(3 + 4x) = 0 has no solutions."

What I don't understand is that it is obvious eq1 and eq2 have no solutions, so how can I possibly show that any roots of eq1 are also roots of eq2?

Thank you.
• September 3rd 2011, 06:05 AM
abhishekkgp
Re: Proving there are no roots to a quadratic equation
Quote:

Originally Posted by mabentley
Hello everyone,

I understand perfectly how to solve quadratic equations, but have come across a question which somewhat confuses me. I hope someone can help:

"Show that any root of the equation 5 + x - sqrt(3 + 4x) = 0 is also a root of the equation x^2 + 6x + 22 = 0. Hence show that the equation 5 + x - sqrt(3 + 4x) = 0 has no solutions."

What I don't understand is that it is obvious eq1 and eq2 have no solutions, so how can I possibly show that any roots of eq1 are also roots of eq2?

Thank you.

$5+x-\sqrt{3+4x} \Rightarrow 5+x=\sqrt{3+4x}$. Square both the sides. What do you get?
• September 3rd 2011, 06:51 AM
mabentley
Re: Proving there are no roots to a quadratic equation

I can see that by squaring the original formula I get a quadratic:

x^2 - 4x + 22 = 0

But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...
• September 3rd 2011, 07:00 AM
Plato
Re: Proving there are no roots to a quadratic equation
Quote:

Originally Posted by mabentley
I can see that by squaring the original formula I get a quadratic:
x^2 - 4x + 22 = 0
But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...

Frankly, I find this to be a very odd question.
I think whoever set the question has a very specific answer in mind.
• September 3rd 2011, 07:11 AM
mabentley
Re: Proving there are no roots to a quadratic equation
Thanks, I'm glad someone agrees. I'm looking back over some A-Level maths (not done any in 7 years) and found this question in the 'past papers' part of an old book. I'm generally finding the questions quite easy, but I'm struggling to see what is wanted from me for this question. I'll give it a miss unless anyone knows better...
• September 3rd 2011, 07:27 AM
abhishekkgp
Re: Proving there are no roots to a quadratic equation
Quote:

Originally Posted by mabentley

I can see that by squaring the original formula I get a quadratic:

x^2 - 4x + 22 = 0

But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...

I think you committed a mistake there.
what i did was
$5+x=\sqrt{3+4x} \Rightarrow (x+5)^2=3+4x \Rightarrow x^2+25+10x=3+4x \Rightarrow x^2+(10-4)x + 25-3=0 \Rightarrow x^2+6x+22=0$.
Now do you get it?
• September 3rd 2011, 07:28 AM
Prove It
Re: Proving there are no roots to a quadratic equation
Quote:

Originally Posted by mabentley

I can see that by squaring the original formula I get a quadratic:

x^2 - 4x + 22 = 0

But the question asks, "Show that any root of this equation...". I can't see that this equation has any roots...

I don't get that...

\displaystyle \begin{align*} 5 + x &= \sqrt{3 + 4x} \\ (5 + x)^2 &= \left(\sqrt{3 + 4x}\right)^2 \\ 25 + 10x + x^2 &= 3 + 4x \\ x^2 + 6x + 22 &= 0 \end{align*}

which is the quadratic you were given...
• September 3rd 2011, 07:45 AM
mabentley
Re: Proving there are no roots to a quadratic equation
Quote:

Originally Posted by abhishekkgp
I think you committed a mistake there.
what i did was
$5+x=\sqrt{3+4x} \Rightarrow (x+5)^2=3+4x \Rightarrow x^2+25+10x=3+4x \Rightarrow x^2+(10-4)x + 25-3=0 \Rightarrow x^2+6x+22=0$.
Now do you get it?

Oops, you are right. That was very sloppy of me - I squared each individual term rather than squaring both sides. It makes complete sense now.