Question

find the zeroes of quadratic polynomial y2 +4 root 3y -15 ​

Answer 1

$\bold{\huge{\fbox{\color{Red}{Solution}}}}$

$= > {y}^{2} + 4 \sqrt{3} y - 15 = 0 \\ = > {y}^{2} + 5 \sqrt{3} y - \sqrt{3} y - 15 = 0 \\ = > y(y + 5 \sqrt{3} ) - \sqrt{3} (y + 5 \sqrt{3)} = 0 \\ = > (y - \sqrt{3} )(y + 5 \sqrt{3} ) = 0 \\ = > y = \sqrt{3} or \: y = - 5 \sqrt{3}$

Answer 2

$\huge{\underline{\underline{\mathtt{\purple{Question}}}}}$

• find the zeros of quadratic polynomial

y2 +4√3y -15

$\huge{\underline{\underline{\mathtt{\purple{Solution}}}}}$

We need to factorise to find out the zeros of this quadratic polynomial

Solving this factorise by splitting middle term

Hence,

$\large\implies\sf y^2+4\sqrt{3y}-15=0$

$\large\implies\sf y^2+5\sqrt{3y}-\sqrt{5y}-15=0$

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

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

Either

y = √3

or

y = -5√3

√3 and -5√3 are the zeros of quadratic polynomial.

$\huge{\underline{\underline{\mathtt{\purple{Verification}}}}}$

We observe that, sum of its zeros

= -5√3 + √3 = √3(-5+1)= -4√3

= $\sf \large\frac{Coefficient\:of\:y}{Coefficient\:of\:y^2}$

Now, Product of its Zeros

= -5√3 × √3 = -5 × 3 = -15

= $\sf \large\frac{Constant\:term}{Coefficient\:of\:y^2}$