Question

Find the value of K, if the pair of linear equation 2x^{2} _5x+3=0 and hence write the nature of the roots
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Answer 1

Answer:

1, 3/2

Real and distinct.

Step-by-step explanation:

Sorry, but I think it is not k but x.

2x²-5x+3=0

2x²-2x-3x+3=0

2x(x-1)-3(x-1)=0

(x-1)(2x-3)=0

x-1=0 or 2x-3=0

x=1 or x=3/2

Nature of roots are real and distinct. Discriminate is greater than zero.

You can also find value of x by quadratic formula and completing square method.

Pls do me a favour. If my answer is wrong report it I don't want others to get wrong knowledge just because of me.

Answer 2

Question :- Find the value of x, if the pair of linear equation 2x² - 5x + 3 = 0 and hence write the nature of the roots ?

Concept used :-

If A•x² + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by :- D = B² - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots..

and,

Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

• x = [ - b ± √D /2a ] where D = Discriminant .

Solution :-

Comparing the given quadratic Equation 2x² - 5x + 3 = 0 , with Ax² + Bx + c = 0, we get,

• A = 2
• B = (-5)
• C = 3

So,

→ D = B² - 4•A•C

→ D = (-5)² - 4*2*3

→ D = 25 - 24

→ D = 1

→ D > 0 .

Therefore, then the given quadratic equation has real and distinct roots.

Now,

→ x = [ - b ± √D /2a ]

Putting values again , and also D = 1 , we get,

→ x = [ - (-5) ± √1 / 2*2 ]

→ x = (5 ± 1) / 4

→ x = (5 + 1)/4 or, (5 - 1)/4

→ x = (6/4) or, (4/4)

→ x = (3/2) or, 1.

Hence, value of x can be {(3/2), 1} and nature of roots are real and distinct .