Advertisements
Advertisements
प्रश्न
Find the roots of the following equation, if they exist, by applying the quadratic formula:
x2 – 4ax – b2 + 4a2 = 0
योग
Advertisements
उत्तर
The given equation is x2 – 4ax – b2 + 4a2 = 0
Comparing it with Ax2 + Bx + C = 0, we get
A = 1, B = –4a and C = –b2 + 4a2
∴ Discriminant, D = B2 – 4AC
= (–4a)2 – 4 × 1 × (–b2 + 4a2)
= 16a2 + 4b2 – 16a2
= 4b2 > 0
So, the given equation has real roots.
Now, `sqrt(D) = sqrt(4b^2) = 2b`
∴ `α = (-B + sqrt(D))/(2A)`
= `(-(-4a) + 2b)/(2 xx 1)`
= `(4a + 2b)/2`
= 2a + b
`β = -(-B - sqrt(D))/(2A)`
= `(-(-4a) - 2b)/(2 xx 1)`
= `(4a - 2b)/2`
= 2a – b
Hence, (2a + b) and (2a – b) are the roots of the given equation.
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
