If zeros of -f(x) = x + 4(square root of x) + 3-are 9 &1 then how come it doesn't equal zero when plugged in?

Answer 1

I think your problem is with the "squareroot of x" part.

In fact, squareroot of a number (positive) gives two answers: one positive and the other negative.

e.g. squareroot of 9 = 3 or -3

Using squareroot of 9 = -3,

f(9) = 9 + 4(-3) + 3 = 0

So, you see 9 is indeed one zero of the function.
Same thing for 1.

Answer 2

staggering At some point has already shown that the only suggestions to this equation are genuine numbers. JB has shown that -12 is a answer. i will coach that -12 is the only answer. because of the fact that's been shown that there are no longer any non-genuine complicated suggestions, then any variables i take advantage of would be assumed to symbolize genuine numbers. --- declare a million. If t is a answer to your equation, then the two t ? -6 or t ? 8. evidence OF declare a million. If t is a answer, then unavoidably t^2 - 2t - 40 8 ? 0 (because of the fact t^2 - 2t - 40 8 is comparable to completely the fee of a few quantity). for this reason (t - 8)(t + 6) ? 0; for this reason t ? -6 or t ? 8. --- declare 2. The inequality x^2 - x - 36 ? 0 is actual if and on condition that the two x ? -5.fifty two (approximately) or x ? 6.fifty two (approximately). evidence OF declare 2. by ability of the quadratic formula and the element theorem, that's the comparable as (x - (a million + ?a hundred forty five) / 2)(x - (a million - ?a hundred forty five) / 2) ? 0 this is actual in basic terms whilst the two x ? (a million - ?a hundred forty five) / 2 or whilst x ? (a million + ?a hundred forty five) / 2; this is, in basic terms whilst x ? -5.fifty two (approximately) or whilst x ? 6.fifty two (approximately). --- declare 3. If t is a answer to your equation, then t^2 - t - 36 ? 0. evidence OF declare 3. assume that t is a answer to your equation. Then, by ability of declare a million, the two t ? -6 or t ? 8. thus, extra frequently, that's actual that the two t ? -5.fifty two or t ? 6.fifty two. thus, by ability of declare 2, that's actual that t^2 - t - 36 ? 0. --- declare 4. If t is a answer to your equation, then ||||||t^2 - t - a million| - 3| - 5| - 7| - 9| - 11| = t^2 - t - 36. evidence OF declare 4. assume that t is a answer to your equation. Then, by ability of declare 3, t^2 - t - 36 ? 0. Then actually t^2 - t - ok ? 0 for any style ok < 36. thus, we've ||||||t^2 - t - a million| - 3| - 5| - 7| - 9| - 11| = |||||t^2 - t - a million - 3| - 5| - 7| - 9| - 11| = |||||t^2 - t - 4| - 5| - 7| - 9| - 11| = ||||t^2 - t - 4 - 5| - 7| - 9| - 11| = ||||t^2 - t - 9| - 7| - 9| - 11| = ... = t^2 - t - a million - 3 - 5 - 7 - 9 - 11 = t^2 - t - 36. --- declare 5. If t is a answer to your equation, then t = -12. evidence OF declare 5. assume that t is a answer to your equation. Then ||||||t^2 - t - a million| - 3| - 5| - 7| - 9| - 11| = t^2 - 2t - 40 8. by ability of declare 4, the left element is comparable to t^2 - t - 36, so t^2 - t - 36 = t^2 - 2t - 40 8 -t - 36 = -2t - 40 8 t - 36 = -40 8 t = -12.

Answer 3

There are no zeroes for this funtcion. The domain (because of the square root of x term) is x>=0. When x is >= 0, all these terms are positive, so f(x)>=3 no matter what x is. f(x)=/=0 ever. The values that you solved for (factor as (sqrt(x)+3)*(sqrt(x)+1)=0 then solve) are extraneous solutions. This sometimes happens when you square both sides of an equality.

Example:

1 = -1 false
(1)^2 = (-1)^2 square both sides
1=1 true

Squaring made the false equation into a true one. When you square to solve, you *need* to check your answers.

Answer 4

because those are not the zeros?

Perhaps you've mis-transcribed the question, but let's walk through this.

0 = x + 4*sqrt(x) +3
3 = x + 4*sqrt(x)
3-x = 4*sqrt(x)

Let's get rid of the sqrt by squaring both sides....

9 -6x + x^2 = 16x

9 - 22x + x^2 = 0

now, remembering the quadratic equation

( b^2 +/- sqrt(b^2-4ac) ) / 2a

22^2 +/- sqrt(22^2 - 4*1*9) / 2

gives 231.416995

and 252.583005

Answer 5

f(x)=x+4x^(.5)+3

Doesn't have any real root since always f(x)>0. x must be positive real number in order x^(.5) be defined, so there is no real root

Answer 6

how do you know they are zeros?...id make sure you have your function right