Question

solve 6 plss i'll mark as brainliest​

Answer 1

Question

The Equation 4x²+(k-2)x+1=0 has no real roots  Find the range of value of k

Solution

Given

4x²+(k-2)x+1=0 has no real roots

To find

The range of value of k

Now take

4x²+(k-2)x+1=0

so compare with

ax² + bx + c + 0

we get

a=4 ,b = (k-2)  and c= 1

For No real roots

Discriminant(D) < 0 , where D = b²-4ac < 0

By putting the value on formula , we get

(k-2)²-4×4×1<0

(k-2)²-16<0

(k-2)²<16

k-2<±4

k<±4+2

When +4

k<4+2

k<6

When -4

k<-4+2

k<-2

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\: the \:Question}}}}$

Here in this question, concept of factorisation is used. We have given an equation and we have to find for value of k will the given equation has no roots.

Roots of any quadratic equation can be classified by value of discriminant as::

$\boxed{\begin{array}{c |c }\bf Discriminant &\bf Nature&&&\sf Positive&\sf Two \:distinct \:roots&\sf Negative &\sf No\:real\:roots &\sf 0&\sf Two\: equal\: roots \end{array}}$

It means that value of discriminant should be less than 0 if the equation has no real roots.

» Given equation::

• 4x²+(k-2)x+1=0

» Here value of::

• a=4

• b=(k-2)

• c=1

» Applying formula for discriminant::

⇒ D = b²-4ac

⇒ 0 > (k-2)²-4(4)(1)

⇒ 0 > (k-2)²-16

» Applying identity:-

(A-B)²=A²+B²-2AB, where A=k and B=2

⇒ 0 > k²+2²-2(k)(2)-16

⇒ 0 > k²+2²-4k-16

⇒ 0 > k²+4-4k-16

⇒ 0 > k²-4k-12

Now factorisation::

⇒ 0 > k²-6k+2k-12

⇒ 0 > k(k-6)+2(k-6)

⇒ 0 > (k-6)(k+2)

⇒ 0 > k=6,-2

So the value of k will be::

k ≥ -2

k≤ 6