Question

2.
The condition that sin 0,cos O may be the roots of ax+ bx + c = 0 is
a) a (a + 2b) = c2 b) a (a +2c) = b c) b(b + 2c) =a² d) b(b+2a)=c²​

Answer 1

★Given :-

sinθ , cosθ may be the roots of ax² + bx + c = 0

★ To find :-

Satisfied condition

★ SOLUTION :-

As they given,

sinθ , cosθ may be the roots of ax² + bx + c = 0

Since,

If α, β are the roots We know ,

$\alpha + \beta = \dfrac{ - b}{a}$

$\alpha \beta = \dfrac{c}{a}$

According to the Question,

$\blue{sin \theta + cos \theta = \dfrac{ - b}{a} - - eq \: 1}$

$\blue{sin \theta \: cos \theta \: = \dfrac{c}{a} - - eq2}$

We shall do squaring on both sides for eq 1

$(sin \theta + cos \theta) {}^{2} = \bigg(\dfrac{ - b}{a} \bigg) {}^{2}$

Expanding the L.H.S equation by (a+b)² = a² + 2ab + b²

$sin {}^{2} \theta + cos {}^{2} \theta + 2sin \theta \: cos \theta \: = \dfrac{b {}^{2} }{a {}^{2} }$

From, Trigonometric identities We know that,

sin²A + cos²A = 1

$1 + 2sin \theta \: cos \theta \: = \dfrac{b {}^{2} }{a {}^{2} }$

We know that ,

2sinθ cosθ = c/a

$1 + \dfrac{2c}{a} = \dfrac{b {}^{2} }{a {}^{2} }$

Take L.C.M in the L.H.S

$\dfrac{a + 2c}{a} = \dfrac{b {}^{2} }{a {}^{2} }$

$a + 2c = \dfrac{b {}^{2} }{a}$

Do cross multiplication

$\blue {a(a + 2c) = b {}^{2} }$

${\boxed{So,\: the \:Required\: Condition\: is \: a(a+2c) = b^2}}$

★Required Answer :-

a( a+ 2c) = b² [B]

Answer 2

★Given :-

sinθ , cosθ may be the roots of ax² + bx + c = 0

★ To find :-

Satisfied condition

★ SOLUTION :-

As they given,

sinθ , cosθ may be the roots of ax² + bx + c = 0

Since,

If α, β are the roots We know ,

$\alpha + \beta = \dfrac{ - b}{a}$

$\alpha \beta = \dfrac{c}{a}$

According to the Question,

$\blue{sin \theta + cos \theta = \dfrac{ - b}{a} - - eq \: 1}$

$\blue{sin \theta \: cos \theta \: = \dfrac{c}{a} - - eq2}$

We shall do squaring on both sides for eq 1

$(sin \theta + cos \theta) {}^{2} = \bigg(\dfrac{ - b}{a} \bigg) {}^{2}$

Expanding the L.H.S equation by (a+b)² = a² + 2ab + b²

$sin {}^{2} \theta + cos {}^{2} \theta + 2sin \theta \: cos \theta \: = \dfrac{b {}^{2} }{a {}^{2} }$

From, Trigonometric identities We know that,

sin²A + cos²A = 1

$1 + 2sin \theta \: cos \theta \: = \dfrac{b {}^{2} }{a {}^{2} }$

We know that ,

2sinθ cosθ = c/a

$1 + \dfrac{2c}{a} = \dfrac{b {}^{2} }{a {}^{2} }$

Take L.C.M in the L.H.S

$\dfrac{a + 2c}{a} = \dfrac{b {}^{2} }{a {}^{2} }$

$a + 2c = \dfrac{b {}^{2} }{a}$

Do cross multiplication

$\blue {a(a + 2c) = b {}^{2} }$

${\boxed{So,\: the \:Required\: Condition\: is \: a(a+2c) = b^2}}$

★Required Answer :-

a( a+ 2c) = b² [B]