Question

17.
It roots of equation x2 +x+r=0 are a 'and'B'and a3 + b3 = -6. Find the value'r'?
(a) = -5/3
(b) =7/3
(C)= -4/3
(d) 1
​

Answer 1

Answer:

square root is 54/74 change it and4735/726

Answer 2

Value of r is -5/3.

Option 'a' is correct.

Given:

• Roots of quadratic equation ${x}^{2} + x + r = 0$ are 'a' and 'b'.
• ${a}^{3} + {b}^{3} = - 6$

To find:

• The value of r is
• (a) = -5/3
• (b) =7/3
• (c)= -4/3
• (d) 1

Solution:

Concept to be used:

1) If $\alpha \: and \: \beta$are roots of quadratic equation $\bf l {x}^{2} + mx + n = 0$ then

$\boxed{ \alpha + \beta = \frac{ - m}{l}}$

and

$\boxed{\alpha \beta = \frac{n}{l}}$

2) $\bf ( {x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

Step 1:

Write the equations from relationship of zeros and coefficients of quadratic equation as discussed above.

As, quadratic equation is ${x}^{2} +x + r = 0$

Coefficients are $\bf l = 1, \: m = 1, and \: n = r$

$a + b = \frac{ - 1}{1}$

or

$\bf a + b = - 1...eq1$

and

$\bf ab = r...eq2$

Step 2:

Find the value of r.

Take cube of eq1.

Apply identity in LHS.

$( {a + b})^{3} = ( { - 1)}^{3}$

or

${a}^{3} + {b}^{3} + 3ab(a + b) = - 1$

Put the given value and values from eq1 and eq2.

$- 6 + 3r( - 1) = - 1$

$- 3r = - 1 + 6$

$\bf r = \frac{ - 5}{3}$

Thus,

Value of r is -5/3.

Option 'a' is correct.

Learn more:

1) if two roots of 2x^2+bx+c=0 are reciprocal of each other then find the value of c

https://brainly.in/question/5471682

2) if sintheta and sectheta (0<theta<90) are the roots of the equation √3x^2+kx+3=0.find k

https://brainly.in/question/21732428

#SPJ3