V1-v2-4 and v21-v22-16- Find value of v1 and v2-

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Standard X

Mathematics

Solving a Quadratic Equation by Completion of Squares Method

v1+v2=4 and ...

Question

$v_{1} + v_{2} = 4$ and $v_{1}^{2} + v_{2}^{2} = 16$ . Find value of $v_{1}$ and $v_{2}$ .

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Solution

$v_{1} + v_{2} = 4$
$v_{1} = 4 - v_{2}$

substitute $v_{1} = 4 - v_{2}$ in the second equation.

${v_{1}}^{2} + {v_{2}}^{2} = 16$

${4 - v_{2}}^{2} + {v_{2}}^{2} = 16$

$16 + {v_{2}}^{2} - 8 v_{2} + {v_{2}}^{2} = 16$

${v_{2}}^{2} - 8 v_{2} + {v_{2}}^{2} = 0$

$2 {v_{2}}^{2} - 8 v_{2} = 0$

${v_{2}}^{2} - 4 v_{2} = 0$

$v_{2} (v_{2} - 4) = 0$

$v_{2} - 4 = 0$

$v_{2} = 4$

now, substitute $v_{2} = 4$ in first equation.

$v_{1} + 4 = 4$

$v_{1} = 0$

so, $v_{1} = 0$ and $v_{2} = 4$ or $v_{1} = 4$ and $v_{2} = 0$

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v1+v2=4 and v21+v22=16. Find value of v1 and v2.

v1+v2=4v1=4−v2substitute v1=4−v2 in the second equation.v12+v22=164−v22+v22=1616+v22−8v2+v22=16v22−8v2+v22=02v22−8v2=0v22−4v2=0v2(v2−4)=0v2−4=0v2=4now, substitute v2=4 in first equation.v1+4=4v1=0so, v1=0 and v2=4 or v1=4 and v2=0

$v_{1} + v_{2} = 4$ and $v_{1}^{2} + v_{2}^{2} = 16$ . Find value of $v_{1}$ and $v_{2}$ .