Question

Check whether the equation 5x² – 6x – 2 = 0 has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

D= b²-4ac is called discriminant.


The nature of roots depend upon the value of the discriminant D. Since D can be zero, positive or negative.


When D>0


If D= b²-4ac >0, then


x= -b+√D/2a &  -b-√D/2a


So, the quadratic equation has two distinct real roots.


[ SOLUTION IS IN THE ATTACHMENT]



Verification:


5x²-6x-2=0


5{(3+√19)/5)}² - 6 (3+√19)/5 -2=0


5(9+6√19+19/25) - (18+6√19)/5 -2=0


(9+6√19+19)/5 - (18+6√19)/5 -2=0


(9+6√19+19)/5 - (18+6√19)/5 -2=0


(9+6√19+19)/5 - 18-6√19/5 -2=0


(9 - 18+19+6√19-6√19)/5  -2=0


(-9+19)/5 -2=0


(10/5)-2=0


2-2= 0


0= 0


L.H.S = R.H.S


Similarly we can prove that


5{(3-√19)/5)}² - 6 (3-√19)/5 -2= 0

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Discriminant = b² - 4ac

Here, a = 5, b = (- 6), c = (- 2)

Then, b² - 4ac = (- 6)² - 4 × 5 × (- 2)

= 36 + 40 = 76 > 0

So the equation has real roots and two distinct roots,

Again, 5x² - 5x = 2 (dividing both sides by 5)

⇒ x² - 6/5x + 9/25 = 2/5 + 9/5

On adding square of half of coefficient of x

⇒ x² - 6/5x + 9/25 = 2/5 + 9/25

⇒ x - 3/5 = ± √19/5

⇒ x = 3 + √19/5 or 3 - √19/5

Verification :-

= 5[3 + √19/5]² - 6[3 + √19/5] - 2

= 9 + 6√19 + 19/5 - (18 + 6√19/5) - 2

= 28 + 6√19/5 - 18 + 6√19/5 - 2

= 28 + 6√19 - 18 - 6√19 - 10/5 = 0

Similarly,

5[3 + √19/5]² - 6[3 + √19/5] - 2 = 0

Hence Verified.