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Elimination Method of Finding Solution of a Pair of Linear Equations

z2+4 z-7=0

Question

z² + 4z – 7 = 0

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Solution

Given: z² + 4z – 7 = 0
=> z²+ 4z = 7 … (1)
We make L.H.S. a perfect square by using
Third term $= {(\frac{1}{2} coefficient of z)}^{2} = {(\frac{1}{2} \times 4)}^{2} = {(2)}^{2} = 4$
On adding 4 to both sides of equation (1), we get:
z² + 4z + 4 = 7 + 4
=> z² + 2 × z × 2 + (2)²= 11
=> (z + 2)² = $(\sqrt{11})^{2}$
On taking square root of both sides, we get:
z + 2 = $\pm$ $\sqrt{11}$

=> z = $-$ $2 \pm \sqrt{11}$
Thus, z = $- 2 + \sqrt{11}, - 2 - \sqrt{11}$

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