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Question
If sin α and cos α are the roots of the equations ax2 + bx + c = 0, then b2 =
Options
a2 − 2ac
a2 + 2ac
a2 − ac
a2 + ac
Solution
The given quadric equation is `ax^2 + bx + c = 0`, and `sin alpha and cos beta` are roots of given equation.
And, a = a,b = b and, c = c
Then, as we know that sum of the roots
`sin alpha + cos beta - (-b)/a`…. (1)
And the product of the roots
`sin alpha .cos beta =c/a`…. (2)
Squaring both sides of equation (1) we get
`(sin alpha + cos beta)^2 = ((-b)/a)^2`
`sin^2 alpha + cos^2 beta + 2 sin alpha cos beta = b^2/a^2`
Putting the value of `sin^2 alpha + cos^2 beta = 1`, we get
`1 + 2 sin alpha cos beta = b^2/a^2`
`a^2 (1+2 sin alpha cos beta) = b^2`
Putting the value of`sin alpha.cos beta = c/a` , we get
`a^2 (1 + 2 c/a) = b^2`
`a^2 ((a+2c)/a) = b^2`
`a^2 + 2ac =b^2`
Therefore, the value of `b^2 = a^2 + 2ac`.
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