Question

If x be real, then the ordinate of vertex of graph formed by quadratic equation 3x² - 10x + 15 = 0, is​

Answer 1

Solving,

3x² - 10x + 15 = 0,

$x_{1,2}=\frac{-(-10) \pm \sqrt{(-10)^{2}-4 \cdot 3 \cdot 15}}{2 \cdot 3}$

$Separate the solutions x_{1}=\frac{-(-10)+4 \sqrt{5} i}{2 \cdot 3}, x_{2}=\frac{-(-10)-4 \sqrt{5} i}{2 \cdot 3}$

$x=\frac{5}{3}+i \frac{2 \sqrt{5}}{3}, x=\frac{5}{3}-i \frac{2 \sqrt{5}}{3}$

Answer 2

Given :  3x² - 10x + 15 = 0

To Find : the ordinate of vertex of graph formed by quadratic equation

Solution:

3x² - 10x + 15 = 0

if x is real then  the ordinate of vertex of graph formed by quadratic equation .

There will be no vertex  but it will be     lines  parallel to x axis as this is an equation .    

also  as (-10)² - 4(3)(15)  = 100 - 240 = -140 < 0 hence no real solution exist.

But if Question is  quadratic polynomial  3x² - 10x + 15

f(x) =  3x² - 10x + 15

f'(x) = 6x - 10

f'(x) = 0

=> x = 10/6 = 5/3

put x = 5/3

f(5/3)  =  3(5/3)² - 10(5/3) + 15

= 25/3 - 50/3 + 15

= 20/3

Then Vertex will be ( 5/3 , 20/3)

abscissa = 5/3

ordinate = 20/3

the ordinate of vertex of graph formed by quadratic polynomial  3x² - 10x + 15    will be 20/3

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