Find the zeros of the following quadratic pol | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the zeros of $p(x)$,we set $p(x) = 0$.$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$We split the middle term $-8x$ into $-6x - 2x$ because $(\sqr

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$,$2 \sqrt{3}$

Vedclass · Answer

(A) To find the zeros of $p(x)$,we set $p(x) = 0$.$\sqrt{3} x^{2} - 8 x + 4 \sqrt{3} = 0$We split the middle term $-8x$ into $-6x - 2x$ because $(\sqrt{3}) 	imes (4\sqrt{3}) = 12$,and we need two numbers whose product is $12$ and sum is $-8$.$\sqrt{3} x^{2} - 6x - 2x + 4 \sqrt{3} = 0$$\sqrt{3} x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$$(x - 2\sqrt{3})(\sqrt{3} x - 2) = 0$Setting each factor to zero:$x - 2\sqrt{3} = 0 \implies x = 2\sqrt{3}$$\sqrt{3} x - 2 = 0 \implies x = \frac{2}{\sqrt{3}}$Thus,the zeros are $\frac{2}{\sqrt{3}}$ and $2\sqrt{3}$.

Find the zeros of the following quadratic polynomial: $p(x)=\sqrt{3} x^{2}-8 x+4 \sqrt{3}$

Similar Questions

The product of the zeros of the polynomial $kx^{3}-6x^{2}+11x-6$ is $4$. Find the value of $k$.

The zero of $p(x) = -x^2 + 2x - 1$ is..........

State the degree of the given polynomial: $p(x) = 5x^{100} - (x^{10})^{20} + 3(x^2)^{25} + 2x^{25} - 7$

Draw the graph of $p(x) = 2x - 5$ and find the zeros of this polynomial.

Which of the following is a cubic polynomial?

Vedclass Test Series

Exam Paper Generator

Online Exam Module