Question

fins the zeros of the quadratic polynomial 15y²-10root6y+10​

Answer 1

Given:

The quadratic polynomial 15y² - 10√6y + 10.

To Find:

The zeros of the given polynomial.

Solution:

The zeros of the quadratic polynomial are those values, when substituted into variables result in zero. In order to find the zeros of the given polynomial, we will first have to split the middle-term of it.

$\sf{\longrightarrow\:15y^2-10\sqrt6y+10=0}$

This can be represented as:

$\sf{\longrightarrow\:5(3y^2-2\sqrt6y+2)=0}$

Moving 5 to the RHS:

$\sf{\longrightarrow\:3y^2-2\sqrt6y+2=0}$

In order to split the middle value into two, the product of the resultant values must be equal to the product of the extreme terms and the sum of the resultant values must be equal to the middle term.

✳ Product = 6y²

✳ Sum = -2√6y

The resultant terms would be -√6y and -√6y.

Splitting:

$\sf{\longrightarrow\:3y^2-\sqrt6y-\sqrt6y+2=0}$

Factorizing:

$\sf{\longrightarrow\:\sqrt3y(\sqrt3y-\sqrt2)-\sqrt2(\sqrt3y-\sqrt2)=0}$

This can be written as:

$\sf{\longrightarrow\:(\sqrt3y-\sqrt2)(\sqrt3y-\sqrt2)=0}$

The values of y are:

$\bold{y=\dfrac{\sqrt2}{\sqrt3},\dfrac{\sqrt2}{\sqrt3}}$

The zeros of the polynomial are √2/√3 and √2/√3 and they are equal.

Answer 2

Answer:

$\sf{Zeros \; of \; the \; polynomial \; are \; \sqrt{\dfrac{2}{3}} \; and \; \sqrt{\dfrac{2}{3}} \; .}$

Explanation:

Given:

$\sf{15y^{2} - 10\sqrt{6}y + 10}$

To find:

Zeros of the given quadratic polynomial.

Solution:

Let,

$\sf{15y^{2} - 10\sqrt{6}y + 10 = 0}$

Factorizing,

$\implies \sf{15y^{2} - 5\sqrt{6}y - 5\sqrt{6}y + 10 = 0}$

$\implies \sf{5\sqrt{3} y(\sqrt{3}y - \sqrt{2}) - 5\sqrt{2}(\sqrt{3}y - \sqrt{2}) = 0}$

$\implies \sf{(\sqrt{3}y - \sqrt{2})(5\sqrt{3} y - 5\sqrt{2}) = 0}$

$\implies \sf{5(\sqrt{3}y - \sqrt{2})(\sqrt{3} y - \sqrt{2}) = 0}$

$\implies \sf{(\sqrt{3}y - \sqrt{2}) = 0 \;\;\;\; or \;\;\;\; (\sqrt{3}y - \sqrt{2}) = 0 }$

$\implies \sf{\sqrt{3}y = \sqrt{2} \;\;\;\; or \;\;\;\; \sqrt{3}y= \sqrt{2} }$

$\implies \sf{y = \dfrac{\sqrt{2}}{\sqrt{3}} \;\;\;\; or \;\;\;\; y = \dfrac{\sqrt{2}}{\sqrt{3}} }$

$\implies \sf{y = \sqrt{\dfrac{2}{3}} \;\;\;\; or \;\;\;\; y = \sqrt{\dfrac{2}{3}} }$

Hence,

$\sf{Zeros \; of \; the \; polynomial \; are \; \sqrt{\dfrac{2}{3}} \; and \; \sqrt{\dfrac{2}{3}} \; .}$