Question

11)
The roots of the equation 5x2-7x+1 are
a. real and equal
b. real roots does not exists
C. real and distinct
d. none of the above​

Answer 1

Answer:-

➥⠀5x² - 7x + 1 has two equal roots.

$\large{\underline{\bf{\blue{Explanation:-}}}}$

➜⠀ Discriminant:-

For the equation ax² + bx + c ,the expression D = (b² - 4ac) is called Discriminant.

D >0 then the equation has two real roots.

D = 0 then the equation has two real and equal roots.

D<0 then the equation has two imaginary roots .

$\large{\underline{\bf{\green{Given:-}}}}$

✰ p(x) = 5x² - 7x + 1

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the nature of roots.

$\huge{\underline{\bf{\red{Solution:-}}}}$

✰ D = ( b² - 4ac )

➜ a = 5

➜ b = -7

➜ c = 1

putting these values in formula.

➩⠀⠀⠀⠀⠀(-7)²- 4×5×1

➩⠀⠀⠀⠀⠀49 - 20

➩⠀⠀⠀⠀⠀29

So , b² - 4ac > 0

SO the given equation have two real roots .

━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\bold\blue{Question}$

The roots of the equation $\bold{5x^{2}-7x+1=0}$ are

a. real and equal

b. real roots does not exists

c. real and distinct

d. none of the above

$\bold\red{\underline{\underline{Answer:}}}$

Correct option is

c.real and distinct

$\bold\orange{Given:}$

The given quadratic equation is

$\bold{5x^{2}-7x+1=0}$

$\bold\pink{To \ find:}$

Nature of the roots.

$\bold\purple{Explanation}$

The nature of root is determined by comparing Discriminant(∆) with zero.

If ∆>0,roots are real and distinct;

∆<0,roots are unreal;

∆=0, roots are real and equal.

$\bold\green{\underline{\underline{Solution}}}$

The given quadratic equation is

$\bold{5x^{2}-7x+1=0}$

Comparing with the general form of quadratic equation

$\bold{ax^{2}+bx+c=0}$

a=5,b=-7 and c=1

Delta(∆)=$\bold{b^{2}-4ac}$

∆=$\bold{-7^{2}-4(5)(1)}$

∆=49-20

∆=29

But,∆>0

Therefore,

$\bold\purple{Roots \ are \ real \ and \ distinct.}$