Question

Find the value of m for which the roots of quadratic equation 4x^2-5mx+1=0 are real and equal​​

Answer 1

Question:

Find the value of m for which the roots of quadratic equation $\sf{{4x}^{2}-5mx+1=0}$ are real and equal.

Given:

• Quadratic equation= $\sf{{4x}^{2}-5mx+1=0}$
• The roots of equation are real and equal.

To Find:

• Value of m.

Formula to be used:

$\sf{D={b}^{2}-4ac}$

Solution:

The given equation is $\sf{{4x}^{2}-5mx+1=0}$. Here,

a=4, b=-5m and c= 1

Substituting all the values in formula. We get,

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

$\rightarrow\mathsf{D={-5m}^{2}-4(4)(1)}$

$\rightarrow\mathsf{D=10m-16}$

The given equation will have real and equal roots, if

$\mapsto\mathsf{D=0}$

$\therefore\mathsf{10m-16=0}$

$\dashrightarrow\mathsf{10m=16}$

$\dashrightarrow\mathsf{m=\large\frac{16}{10}}$

$\dashrightarrow\mathsf{m=\large\frac{8}{5}}$

More to Know:

– Nature of the roots of quadratic equation ax²+bx+c=0 , a≠0 depends upon the value of D=b²-4ac, which is known as the Discriminate of quadratic equation.

– The quadratic equation ax²+bx+c=0, a≠0 has:

1) Two distinct real roots, if D=b²-4ac >0

2) Two equal roots (coincident real roots), if D=b²-4ac=0

3) No real roots, if D=b²-4ac< 0

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Answer 2

${\huge{ \boxed{\underline{\colorbox{pink}{Answer}}}}}$

To Solve:

• Find the value of m for which the roots of quadratic equation 4x² - 5mx + 1 = 0 are real and equal.

Solⁿ:

• Eqⁿ: 4x² - 5mx + 1 = 0
• Equal and Real roots = 0

Discriminant: b² - 4ac

b = -5

a = 4

c = 1

=> (-5m)² - 4(4)(1) = 0

=> 25m² - 16 = 0

$= > {m}^{2} \: = \frac{16}{25} \\ \\ m = \sqrt{ \frac{16}{25} } \\ \\ { \boxed{ \pink{m = \frac{4}{5}}}}$

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