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