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3 years ago

X^2+4x=15 fghhyyhhhhhhjjrrfgh

1 answer:

8 0

$(*)\qquad(a+b)^2=a^2+2ab+b^2\\\\x^2+4x=15\\\\x^2+2(x)(2)=15\qquad\text{add}\ 2^2\ \text{to both sides}\\\\\underbrace{x^2+2(x)(2)+2^2}_{(*)}=15+2^2\\\\(x+2)^2=15+4\\\\(x+2)^2=19\to x+2=\pm\sqrt{19}\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2-\sqrt{19}\ \vee\ x=-2+\sqrt{19}}$

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3 years ago

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Answer:

1. Ans;

${a}^{7} \times {a}^{12} = {a}^{7 + 12} = {a}^{19}$

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2. Ans;

$\frac{ {h}^{6} }{ {h}^{2} } = {h}^{4}$

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3. Ans;

${e}^{6} \times {e}^{6} = {e}^{6 + 6} = {e}^{12}$

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4.Ans;

$\frac{ {w}^{9} }{ {w} } = {w}^{8}$

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5. Ans;

${g}^{2} \times {h}^{3} \times {g}^{2} \times h = {g}^{2 + 2} \times {h}^{3 + 1} = {g}^{4} {h}^{4}$

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6.Ans;

$\frac{10 {a}^{12} }{ 6{a}^{3} } = \frac{5 {a}^{9} }{3}$

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7.Ans;

$({m}^{4} )^{5} = {m}^{4 \times 5} = {m}^{20}$

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8.Ans;

${(3 {a}^{2} })^{3} = ( {3})^{3} ( {a}^{2} ) ^{3} = 27 {a}^{6}$

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9.Ans;

${( \frac{c}{k} })^{4} = \frac{ {c}^{4} }{ {k}^{4} }$

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10.Ans;

${( \frac{8x}{9} })^{2} = \frac{( {8x})^{2} }{ ({9})^{2} } = \frac{64 {x}^{2} }{81}$

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11.Ans;

$4 {c}^{2} \times {c}^{3} = 4 {c}^{2 + 3} = 4 {c}^{5}$

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12.Ans;

${( - 9 {h}^{4} {k}^{6} })^{2} = {( - 9)}^{2} {( {h}^{4} })^{2} {( {k}^{6} })^{2} = 81 \: \: {h}^{8} {k}^{12}$

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13.Ans;

$( { \frac{2d}{3e} })^{3} = \frac{ {(2d)}^{3} }{ ({3e)}^{3} } = \frac{8 {d}^{3} }{27 {e}^{3} }$

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14. Ans;

${( \frac{4}{ {w}^{5} }) }^{3} = \frac{ ({4)}^{3} }{( {w}^{5} ) ^{3} } = \frac{64}{ {w}^{15} }$

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$\frac{5 {v}^{4} \times 4 {v}^{5} }{2 {v}^{3} } = \frac{20 {v}^{4 + 5} }{2 {v}^{3} } = \frac{20 {v}^{9} }{2 {v}^{3} } = 10 {v}^{6}$

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$( \frac{3 {f}^{4}h}{4 {h}^{9} } ) ^{3} = \frac{( {3})^{3} ( {f}^{4})^{3} ({h})^{3} }{ {(4)}^{3} ( {h}^{9})^{3} } \\ \\ \\ = \frac{27 {f}^{4 \times 3} {h}^{3} }{64 {h}^{9 \times 3} } = \frac{27 {f}^{12} {h}^{3} }{64 {h}^{27} } \\ \\ \\ = \frac{27 {f}^{12} }{64 {h}^{24} }$