What is the solution to y - 8y^(1/2) + 11 = 0 using quadratic formula?

  • Thread starter DeanBH
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In summary, the first part of the question asks to solve the equation x^2 - 8x + 11 = 0 using the quadratic formula, which gives the solutions 4 +/- root5. In the second part, the equation y - 8y^(1/2) + 11 = 0 is given and we are asked to solve it in the form p +/- Q * root5. To do this, we can substitute y = x^2 and use the solutions found in the first part as the values for x, which will give us the solutions for y. This substitution makes the problem easier to solve and is a valid method.
  • #1
DeanBH
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first part of the question is simple :
x^2 - 8x + 11 = 0

solve. using quadratic formula it is 4 +/- root5

second part confuses me, you are given to equation:

y - 8y^(1/2) + 11 = 0

and are told to :

solve this giving answer in form p +/- Q * root5

i have no idea how to do this. the quadratic formula doesn't work on this one and i don't understand how part 1 of this question helps me with this part, can someone explain it to me please.!? thanks
 
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  • #2
Hint: Can you express y in terms of x, such that your second equation becomes your first?
 
  • #3
i don't know how to do that, I've tryed making the equations equal to each other but it doesn't work out right
 
  • #4
Compare the equations term by term.

11=11

next is...
 
  • #5
oh yeah, i can do that because they're both 0, and are of the same form. lewl
 
  • #6
DeanBH said:
i don't know how to do that, I've tryed making the equations equal to each other but it doesn't work out right

Hint: substitution. :smile:
 
  • #7
tiny-tim said:
Hint: substitution. :smile:

i've honastly tried for ages, i still don't know how to do it.

can someone just run me through it. it's not even a question for homework or anything I am just revising and i don't understand this.
 
  • #8
Bigger hint: substitute y = x². :smile:
 
  • #9
tiny-tim said:
Bigger hint: substitute y = x². :smile:


why can i just substitute that.
 
  • #10
DeanBH said:
why can i just substitute that.

eh? :confused: You can substitute anything you like. :smile:

Some substitutions make the problem easier :!) , some substitutions make it harder. :mad:

But all substitutions are valid.

Try it … put y = x² into y - 8√y + 11 = 0, and see what happens! :smile:
 
  • #11
DeanBH said:
why can i just substitute that.
think about it a second:
you've already found that [tex]x^{2} - 8x + 11[/tex]
Now you need to find [tex]y - 8y^{1/2} + 11[/tex]
If you substituted [tex]x^{2}[/tex] = y,
you would have [tex]x^{2} - 8x + 11[/tex]
which you already have the answer to. If x = 4+/-[tex]\sqrt{5}[/tex],
what is [tex]x^{2}[/tex] (i.e. y) going to equal?
 

1. How do I solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula, factoring, or completing the square. It is important to first make sure that the equation is in standard form (ax^2 + bx + c = 0) before attempting to solve it.

2. What is the quadratic formula?

The quadratic formula is a formula that can be used to solve any quadratic equation in the form of ax^2 + bx + c = 0. It is written as x = (-b ± √(b^2 - 4ac)) / 2a.

3. How do I use factoring to solve a quadratic equation?

To use factoring, you need to first factor the quadratic equation into two binomials. Then, set each binomial equal to 0 and solve for the variable. This will give you two solutions for the equation.

4. What is completing the square and how does it help solve a quadratic equation?

Completing the square is a method used to solve a quadratic equation by manipulating the equation into a perfect square trinomial. This makes it easier to solve for the variable by taking the square root of both sides of the equation.

5. What do the solutions of a quadratic equation represent?

The solutions of a quadratic equation represent the x-intercepts of the graph of the equation. They are the values of x where the equation crosses or touches the x-axis.

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