Question

What is the sum product and roots of 2x²-3x=0

Answer 1

Step-by-step explanation:

2x²-3x=0

in the above equation

a = 2

b= -3

c = 0

sum of the roots = -b/a

= -(-3)/2

= 3/2

product of roots = c/a

= 0/2

= 0

roots of 2x²-3x=0

x(2x - 3) = 0

x=0 and 2x-3=0

x =0 and x = 3/2

roots are 0 and 3/2

Answer 2

Given:

An equation, $2x^{2} -3x=0$.

To find:

Sum, product and roots of the equation.

Solution:

If an equation is of the form $ax^{2} +bx+c=0$, then the roots of the equation is given by calculating the discriminant, $D=-b\pm\sqrt{b^{2}-4ac}$.

The roots are given as $x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$, $x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

Sum is given as $b$ and product is given as $c*a$.

Here, we have,  $2x^{2} -3x=0$.

$a=2,b=-3,c=0$

$Sum=-3$

$Product=0$

Roots,

$x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

$x_{1}=\frac{-(-3)+\sqrt{(-3)^{2}-4*2*0}}{2*2}$

$x_{1}=\frac{3+\sqrt{9}}{4}$

$x_{1}=\frac{3+3}{4}=\frac{6}{4}=\frac{3}{2}$

$x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

$x_{2}=\frac{-(-3)-\sqrt{(-3)^{2}-4*2*0}}{2*2}$

$x_{2}=\frac{3-\sqrt{9}}{4}$

$x_{2}=\frac{3-3}{4}=\frac{0}{4}=0$

∴ Roots are $\frac{3}{2},0$

The sum, product and roots of the equation are $-3,0$ and $(\frac{3}{2},0)$ respectively.