Question

For what value of ' k ' the following equation have real and equal roots–2x²-4x+k=0​

Answer 1

Given:-

• Quadratic Equation 2x²-4x+k=0 has equal and real roots

To find:-

• Value of k

Solution:-

For real roots the value of Discriminant is greater or equal than 0.

And for equal roots the value of Discriminant must be equal to Zero

So, We have

$\tt \rightarrow \blue{\bf{D=0}}$

We know the formula of Discriminant

$\tt \rightarrow \green{ D = b^2-4ac }$

So we conclude that

$\tt b^2-4ac=0 \\\\ \tt where \\\\ \tt \star a= 2 \\\\ \tt \star b= -4 \\\\ \tt \star c=k$

Now solving the equation further

$\tt (-4)^2-4(2)(k)=0 \\\\ \tt \Rightarrow 16-8k=0 \\\\ \tt \Rightarrow 16=8k\\\\ \Rightarrow \dfrac{16}{8}=k \\\\ \tt \Rightarrow 2=k$

So, we have value of k as 2

Answer 2

$\sf{\underline{\underline{Question:-}}}$

For what value of ' k ' the following equation have real and equal roots–2x²-4x+k=0.

$\sf{\underline{\underline{TO\:FIND:-}}}$

VALUE OF K =?

$\sf{\underline{\underline{Solution:-}}}$

$\sf{\underline{\underline{USING\: DISCRIMINANT \: FORMULA:-}}}$

$\sf{\fbox{\underline{ d=b^2-4ac}}}$

2x²-4x+k=0.

a = 2

b = -4

c = k

★ BY THE FORMULA ★

$\sf→ d = b^2-4ac=0\\\sf→ d= (-4)^2-4×2×k=0\\\sf→ d= 16-8k=0\\\sf→16=8k \\\sf→ \frac{16}{8}=k\\\sf→ k = 2$

$\sf{\underline{\underline{Hence:-}}}$

Value of k = 2