The roots of the equation of a quadratic function are x=3 and x=5. Find an equation for the quadratic function that contains the point(1,-24).

tabita57i
2021-08-12
Answered

The roots of the equation of a quadratic function are x=3 and x=5. Find an equation for the quadratic function that contains the point(1,-24).

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hesgidiauE

Answered 2021-08-13
Author has **106** answers

The roots are x=3 and x=5

f(x) =a(x-3)(x-5)

It passes through (1,-24)

-24=a(1-3)(1-5)

-24=a(-2)(-4)=8a

$a=-\frac{24}{8}a=-3$

f(x)=-3(x-3)(x-5)

=$-3({x}^{2}-5x-3x+15)$

$f\left(x\right)=-3{x}^{2}+24x-45$

f(x) =a(x-3)(x-5)

It passes through (1,-24)

-24=a(1-3)(1-5)

-24=a(-2)(-4)=8a

f(x)=-3(x-3)(x-5)

=

Jeffrey Jordon

Answered 2022-01-14
Author has **2581** answers

Answer is given below (on video)

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My attempt is as follows:

$(x-d\frac{2k-\sqrt{4{k}^{2}-20}}{10})(x-d\frac{2k+\sqrt{4{k}^{2}-20}}{10})<0$

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$x\in (d\frac{k-\sqrt{{k}^{2}-5}}{5},d\frac{k+\sqrt{{k}^{2}-5}}{5})$

As it is given that it has got only one integral solution, so there must be exactly one integer between$d\frac{k-\sqrt{{k}^{2}-5}}{5}$ and $d\frac{k+\sqrt{{k}^{2}-5}}{5}$ .

Let$x}_{1}=d\frac{k-\sqrt{{k}^{2}-5}}{5$ and $x}_{2}=d\frac{k+\sqrt{{k}^{2}-5}}{5$ , then $\left[{x}_{2}\right]-\left[{x}_{1}\right]=1$ where [] is a greater integer function.

But from here, how to proceed? Please help me in this.

My attempt is as follows:

As it is given that it has got only one integral solution, so there must be exactly one integer between

Let

But from here, how to proceed? Please help me in this.

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