Question

find a quadratic polynomial for which the sum and the product of zeroes are root 5 and 3/4​

Answer 1

a + b = √5

ab = 3/4

So the equation will be

4x^2 + 4✓5x + 3

i hope it would help :)

Answer 2

Answer:

$x^2-\sqrt{5}x+\frac{3}{4}$

Step-by-step explanation:

Let $x_1$ and $x_2$ be the roots of the polynomial $ax^2+bx+c$

We have the formula for sum and product of roots of a quadratic polynomial:

$x_1+x_2 &= \frac{-b+\sqrt{b^2-4ac} }{2a} +\frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-b+\sqrt{b^2-4ac}-b-\sqrt{b^2-4ac} }{2a} = \frac{-2b} {2a}= \frac{-b}{a}$

and:

$x_1x_2= \frac{-b+\sqrt{b^2-4ac} }{2a}. \frac{-b+\sqrt{b^2-4ac} }{2a}= \frac{(-b)^2-(\sqrt{b^2-4ac})^2 }{4a^2}=\frac{b^2-(b^2-4ac)}{4a^2}=\frac{4ac}{4a^2}=\frac{c}{a}$

So you have to solve the following system of equations:

$\left \{ {{\frac{-b}{a}=\sqrt{5} } \atop {\frac{c}{a}=\frac{3}{4}} \right$

Which is the same as:

$\left \{ {{-b=\sqrt{5}a } \atop {c=\frac{3}{4}a} \right$

So: $\left \{ {{b=-\sqrt{5}a } \atop {c=\frac{3}{4}a} \right$

From this we get the polynomial:

$ax^2-\sqrt{5}ax+\frac{3}{4}a$

Here $a$ can take any value (besides 0)

So the final answer is (we can take $a=1$):

$x^2-\sqrt{5}x+\frac{3}{4}$

Note:

You may notice that the sun and product appear in our polynomial.

$x^2-Sx+P$

Where $S$ and $P$ are the sum and productof the roots respectively.

So if you want to find two numbers where their sum is $S$ and product is $P$, you can solve the equation: $x^2-Sx+P=0$

This is known as Vieta's formulas, and  it applies to even higher degree polynomials.

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