**If 2a – b + c = 0, prove that 4a**^{2} – b^{2} + c^{2} + 4ac = 0.

^{2}– b

^{2}+ c

^{2}+ 4ac = 0.

**Answer :**

It is given that

2a – b + c = 0

2a + c = b

By squaring on both sides

(2a + c)^{2 }= b^{2}

Expanding using formula

(2a)^{2} + 2 × 2a × c + c^{2} = b^{2}

By further calculation

4a^{2} + 4ac + c^{2} = b^{2}

So we get

4a^{2} – b^{2} + c^{2} + 4ac = 0

Hence, it is proved.

**More Solutions:**

- Without actually calculating the cubes, find the values of:
- Using suitable identity, find the value of:
- If x – y = 8 and xy = 5, find x2 + y2.
- If x + y = 10 and xy = 21, find 2 (x2 + y2).
- If 2a + 3b = 7 and ab = 2, find 4a2 + 9b2.
- Find the value of 9×2 + 16y2.
- Find the value of 2×2 + 2y2.
- If a2 + b2 = 13 and ab = 6, find
- If a + b = 4 and ab = -12, find
- If p – q = 9 and pq = 36, evaluate