Question

if the sum of the roots of the quadratic equation ax^2+bx+c=0 is equal to the sum of the squares of their reciprocals then a^2/ac+bc/a2​

Answer 1

Correct Question

If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals then b²/ac + bc/a²

Solution

Let the roots of the equation be M and N.

Given that, quadratic equation is ax² + bx + c = 0

Sum of zeros = -b/a

→ M + N = -b/a

Product of zeros = c/a

→ MN = c/a

According to question,

→ M + N = 1/M² + 1/N²

→ M + N = (N² + M²)/M²N²

→ M + N = [(N + M)² - 2MN]/M²N²

Used identity is (a + b)² = a² + b² + 2ab

So, (a + b)² - 2ab = a² + b² + 2ab - 2ab

= a² + b²

→ M + N = [(M + N)² - 2MN]/(MN)²

→ -b/a = [(-b/a)² - 2(c/a)]/(c/a)²

→ -b/a = [(b²/a² - 2c/a)]/(c²/a²)

→ -b/a × c²/a² = b²/a² - 2c/a

→ -bc²/a³ = b²/a² - 2c/a

→ -bc²/a³ - b²/a² = - 2c/a

→ (-bc² - b²a)/a³ = - 2c/a

→ -b(c² + ba)/a³ = -2c/a

→ b(c² + ba)/a³ = 2c/a

→ ba(c² + ba)/a³ = 2c

→ b(c² + ba)/a² = 2c

→ (bc² + b²a)/a² = 2c

→ bc²/a²c + b²a/a²c = 2

→ bc/a² + b²/ac = 2

Answer 2

$\huge \boxed {question}$

If the Roots of a quadratic equation ax²+bx+c = 0 are equal to the sum of squares of their reciprocal then value of

$\frac{ {b}^{2} }{ac} + \frac{bc}{ {a}^{2} }$

Let's assume that the roots of the equation are x and y .

$\huge\boxed{Given}$

$x + y = \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} }$

$\huge\boxed{Solution}$

➲Given that

$x + y = ({ \frac{1}{x} )}^{2} + ({ \frac{1}{y}) }^{2}$

➲So , Taking LCM ·

$x + y = \frac{ {y}^{2} + {x}^{2} }{ {x}^{2} . {y}^{2} }$

Equation 1st ⇑

➲To find the value of x² + y² we will use (x+y)² :-

➲(x+y )² = x² + y² + 2x.y

➲ x² + y² = ( x+y)² - 2x.y

➲ Now putting the value of x² + y² in eq. 1st .

$x + y = \frac{( {x + y)}^{2} - 2x.y }{( {x +.y)}^{2} }$

➲Now we know that Sum of roots is equal to

-b/a

Square of sum of roots is b²/a²

and multiple of roots is c/a .

$\frac{ - b}{a} = \frac{ \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} }{ ({ \frac{c}{a} )}^{2} }$

$\frac{ - b}{a} = \frac{ \frac{ {b}^{2} - 2ac}{a} }{ \frac{ {c}^{2} }{ {a}^{2} } }$

$\frac{ - b}{a} = \frac{ {b}^{2} - 2ac }{ {c}^{2} }$

$\frac{ - b {c}^{2} }{a} = {b}^{2} - 2ac$

$2ac = {b}^{2} + \frac{b {c}^{2} }{a}$

$2 = \frac{ {b}^{2} }{ac} + \frac{b {c}^{2} }{a \times ac}$

$\boxed{2 = \frac{ {b}^{2} }{ac} + \frac{bc}{ {a}^{2} } }$

Answer is 2