Question

1.
Find the roots of the following quadratic equations, ifthey exist
i 2x²+x-4 = 0
ii 4x^2 +4√3x + 3 = 0
iii 5x^2- 7x - 6 = 0
iv. x² +5= 6x​

Answer 1

Step-by-step explanation:

oots( £,β) of any quadratic equation ax² + bx + c = 0 can be found out by …..

£,β = {-b+,- √( b² - 4ac)} / 2a , where a,b,c are real numbers

So, Nature of the roots £,β of any quadratic equation, depends on the expression (b² -4ac) , ie, this expression discriminates the roots of the quadratic equation.

So, we call it Discriminant ( D) = b² - 4ac

So, ‘D’ of 2x² - 5x + 6 =0

= 25 - 48 = - 23

Since it is negative, or D < 0

So, the equation has IMAGINARY & UNEQUAL roots

& roots £,β are a pair of COMPLEX CONJUGATES

Answer 2

Answer:

The roots of quadratic equation are the values of the variable that satisfy the equation. They are also known as the "solutions" or "zeros" of the quadratic equation. For example, the roots of the quadratic equation x2 - 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation. i.e.,

when x = 2, 22 - 7(2) + 10 = 4 - 14 + 10 = 0.

when x = 5, 52 - 7(5) + 10 = 25 - 35 + 10 = 0.

But how to find the roots of a general quadratic equation ax2 + bx + c = 0? Let us try to solve it for x by completing the square.

ax2 + bx = - c

Dividing both sides by 'a',

x2 + (b/a) x = - c/a

Here, the coefficient of x is b/a. Half of it is b/(2a). Its square is b2/4a2. Adding b2/4a2 on both sides,

x2 + (b/a) x + b2/4a2 = (b2/4a2) - (c/a)

[ x + (b/2a) ]2 = (b2 - 4ac) / 4a2 (using (a + b)² formula)

Taking square root on both sides,

x + (b/2a) = ±√ (b² - 4ac) / 4a²

x + (b/2a) = ±√ (b² - 4ac) / 2a

Subtracting b/2a from both sides,

x = (-b/2a) ±√ (b² - 4ac) / 2a (or)

x = (-b ± √ (b² - 4ac) )/2a

This is known as the quadratic formula and it can be used to find any type of roots of a quadratic equation.