Question

If \( \sin\alpha \) and \( \cos\alpha \) are the roots of equation : \( ax^2+bx+c=0 \) then \( b^2 = \)

• A. \(a^2 - 2ac\)
• B. \(a^2 + 2ac\)
• C. \(a^2 - ac\)
• D. \(a^2 + ac\)

Hint:

Given roots \(r_1 = \sin\alpha, r_2 = \cos\alpha\) for \(ax^2+bx+c=0\). 1. Sum of roots: \(\sin\alpha + \cos\alpha = -b/a\). 2. Product of roots: \(\sin\alpha \cos\alpha = c/a\). 3. Square the sum of roots: \((\sin\alpha + \cos\alpha)^2 = (-b/a)^2\). \(\sin^2\alpha + \cos^2\alpha + 2\sin\alpha\cos\alpha = b^2/a^2\). 4. Substitute identity \(\sin^2\alpha + \cos^2\alpha = 1\) and product from step 2: \(1 + 2(c/a) = b^2/a^2\). 5. Multiply by \(a^2\): \(a^2 + 2ac = b^2\).

Answer 1

The Correct Option is B

Concept: For a quadratic equation \( Ax^2+Bx+C=0 \), if the roots are \(r_1\) and \(r_2\), then:
Sum of roots: \(r_1 + r_2 = -B/A\)
Product of roots: \(r_1 r_2 = C/A\) We also use the fundamental trigonometric identity: \(\sin^2\alpha + \cos^2\alpha = 1\). Step 1: Apply Vieta's formulas to the given equation The given equation is \( ax^2+bx+c=0 \). The roots are given as \( \sin\alpha \) and \( \cos\alpha \). Comparing with \(Ax^2+Bx+C=0\), we have \(A=a, B=b, C=c\). Sum of roots: \[ \sin\alpha + \cos\alpha = -\frac{b}{a} \quad \cdots (1) \] Product of roots: \[ \sin\alpha \cos\alpha = \frac{c}{a} \quad \cdots (2) \] Step 2: Use the identity \((x+y)^2 = x^2 + y^2 + 2xy\) We want to find an expression involving \(b^2\). Notice that \(b\) appears in the sum of roots. Let's square the equation (1): \[ (\sin\alpha + \cos\alpha)^2 = \left(-\frac{b}{a}\right)^2 \] Expand the left side: \[ \sin^2\alpha + \cos^2\alpha + 2\sin\alpha\cos\alpha = \frac{b^2}{a^2} \] Step 3: Substitute known identities and expressions We know the trigonometric identity: \( \sin^2\alpha + \cos^2\alpha = 1 \). From equation (2), we know: \( \sin\alpha\cos\alpha = \frac{c}{a} \). Substitute these into the expanded equation from Step 2: \[ 1 + 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} \] Step 4: Solve for \(b^2\) \[ 1 + \frac{2c}{a} = \frac{b^2}{a^2} \] To clear the denominators, multiply the entire equation by \(a^2\): \[ a^2\left(1 + \frac{2c}{a}\right) = a^2\left(\frac{b^2}{a^2}\right) \] \[ a^2(1) + a^2\left(\frac{2c}{a}\right) = b^2 \] \[ a^2 + 2ac = b^2 \] So, \( b^2 = a^2 + 2ac \).