Math, asked by Palandr3576, 8 months ago

Find the roots of quadratic equations √x(x-7) =3√2 by applying quadratic formula

Answers

Answered by anyasood
0

Quadratic equation:

An equation of the form ax2+bx +c = 0 is called a quadratic equation in one variable, where a, b, c are real numbers and a ≠ 0.

Quadratic Formula:

This method is also called as Sridharacharya's rule.

x= –b±√b2 – 4ac/2a

where b2 - 4ac is called the discriminant of the quadratic equation and it is denoted by 'D'.

D= b2 - 4ac

x= -b±√D/2a

Nature of the roots

If D = 0 roots are real and equal , D > 0 roots are real and unequal, D < 0 roots are imaginary.

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Solution:

(i) 2x2 – 7x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get

a = 2, b = -7 and c = 3

By using quadratic formula, we get

x = –b±√b2 – 4ac/2a

⇒ x = 7±√49 – 24/4

⇒ x = 7±√25/4

⇒ x = 7±5/4

⇒ x = 7+5/4 or x = 7-5/4

⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x2 + x – 4 = 0

On comparing this equation with ax2 + bx + c = 0, we get

a = 2, b = 1 and c = -4

By using quadratic formula, we get

x = –b±√b2 – 4ac/2a

⇒x = -1±√1+32/4

⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x2 + 4√3x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

By using quadratic formula, we get

x = –b±√b2 – 4ac/2a

⇒ x = -4√3±√48-48/8

⇒ x = -4√3±0/8

∴ x = √3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

On comparing this equation with ax2 + bx + c = 0, we get

a = 2, b = 1 and c = 4

By using quadratic formula, we get

x = –b±√b2 – 4ac/2a

⇒ x = -1±√1-32/4

⇒ x = -1±√-31/4

The square of a number can never be negative.

∴there is no real solution of this equation.

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Hope this will help you.......

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Answered by arvindsingh52
0

Step-by-step explanation:

√x(x-7) =3√2

squre both side

(x(x-7)) ^2=(3√2) ^2

(x^2)(x^2-14x+49)=18

aage isko simplify krne se ho jayega

please mark me bralient

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