Question

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Answer 1

Step-by-step explanation:

$x + {y}^{2} = 12$

$x = 12 - {y}^{2}$

$put \: value \: of \: x \: in \: equation \: \: {x}^{2} + y = 12$

${(12 - {y}^{2} )}^{2} + y = 12$

$144 - 24 {y}^{2} + { {y}^{4} } \: + y - 12 = 0$

${y}^{4} - 24 {y}^{2} + y + 132 = 0$

${y}^{4} - 16 {y}^{2} - 8 {y}^{2} + y + 132 = 0$

${y}^{2} ( {y}^{2} - 16) - (8 {y}^{2} - y - 132) = 0$

${y}^{2} (y - 4)(y + 4) - (8 {y}^{2} + 32y - 33y - 132) = 0$

${y}^{2} (y - 4)(y + 4) - (8y(y + 4) - 33(y + 4) ) = 0$

${y}^{2} (y - 4)(y + 4) - (8y - 33)(y + 4) = 0$

$(y + 4)[{y}^{2} (y - 4) - (8y - 33)] = 0$

${y}^{2} (y - 4) - (8y - 33) = 0$

${y}^{3} - 4 {y}^{2} - 8y + 33 = 0$

${y}^{3} - 4 {y}^{2} + 3y - 11y + 33 = 0$

$y( {y}^{2} - 4y + 3) - 11(y - 3) = 0$

$y(y - 3)(y - 1) - 11(y - 3) = 0$

$(y - 3)[y(y - 1) - 11] = 0$

$y(y - 1) - 11 = 0$

${y}^{2} - y - 11 = 0$

Comparing with the general form of quadratic equation $a {x}^{2} + bx + c = 0$

a = 1, b = -1 c = -11

by formula method

$y = \frac{ - b + - \sqrt{ {b - 4ac}^{2} } }{2a}$

$y = \frac{1 + - \sqrt{1 - 4 \times 1 \times - 11} }{2}$

$y = \frac{1 + - \sqrt{1 + 44} }{2}$

$y = \frac{1 + - \sqrt{45} }{2}$

$y = \frac{1 + - 3 \sqrt{5} }{2}$

$for \: y = \frac{1 + 3 \sqrt{5} }{2} ,\: {y}^{2} = {( \frac{1 + 3 \sqrt{5} }{2}) }^{2}$

${y}^{2} = \frac{1 + 6 \sqrt{5} + 45}{4} = \frac{46 + 6 \sqrt{5} }{4} = \frac{23 + 3 \sqrt{5} }{2}$

$x = 12 - {y}^{2} = 12 - \frac{23 + 3 \sqrt{5} }{2} = \frac{24 - (23 + 3 \sqrt{5}) }{2}$

$x = \frac{24 - 23 - 3 \sqrt{5} }{2} = \frac{1 - 3 \sqrt{5} }{2}$

Therefore, $for \: y = \frac{1 + 3 \sqrt{5} }{2} \: x = \frac{1 - 3 \sqrt{5} }{2}$

Similarly, $for \: y = \frac{1 - 3 \sqrt{5} }{2} \: x = \frac{1 + 3 \sqrt{5} }{2}$

Hence,

$(x ,\: y) = ( \frac{1 + - 3 \sqrt{5} }{2} , \: \frac{1 - + 3 \sqrt{5} }{2} )$