Question

write in the form of ax^2+bx+c=0 and find a,b,c :5x^2-7=-6x​

Answer 1

Answer:

5x^2-7=-6x

5x^2-7+6x

A=5x^2,B=-7,C=6x

Answer 2

Answer:

We have provided with the equation and it is said to find out the value of a, b and c respectively.

What we will do ?

We will compare the given equation with the general form of quadratic equation and arrange it accordingly. Thereafter , we will get our answer.

Let's do it now :

As we know :

General form of quadratic equation is -

$:\implies\sf ax^{2} + bx + c = 0$ { Where , a ≠ 0 }

Then ,

Let's arrange the given equation :

• 5x² - 7 = - 6x

Transposing - 6 x to the left side , then

Equation :

5 x² + 6x - 7 = 0

Therefore, after comparing it with the general form we get ,

• a = 5

• b = 6

• c = - 7

More to know :

Quadratic formula ,

$:\implies\sf \alpha = \dfrac{- b + \sqrt D}{2a}$

And

$:\implies\sf \beta = \dfrac{- b - \sqrt D}{2a}$

The value of the D is = $\sf b^{2} - 4 ac$ ≥ 0.

If D = $\sf b^{2} - 4 ac$ > 0 , then $\sf \alpha \: and \: \beta$ are real { having real roots }.

So, If

$:\implies \sf \alpha - \beta = ( \dfrac{- b - \sqrt D}{2a}) - ( \dfrac{- b + \sqrt D}{2a} )$

$:\implies\sf \dfrac{ - b^{2} + \sqrt D + b + \sqrt D}{ 2a } = \dfrac{2 \sqrt D}{2a} = \dfrac{ \sqrt D}{a}$

→ α - β ≠ 0 → α ≠ β

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