【二次関数】平方完成の演習問題

平方完成の演習問題

LEVEL1

(1)　$x^2-4x+7$

(2)　$x^2+10x+19$

(3)　$x^2-2x+3$

(4)　$x^2-4x+4$

(5)　$x^2+8x+11$

(6)　$x^2-6x+7$

(7)　$x^2-6x$

(8)　$x^2+12x$

詳しい解説は下の方にあります。

LEVEL2

(1)　$x^2+3x+1$

(2)　$x^2-7x$

(3)　$x^2-5x+4$

(4)　$x^2+x-4$

(5)　$x^2-11x+8$

詳しい解説は下の方にあります。

LEVEL3

(1)　$2x^2+4x-1$

(2)　$3x^2-12x+3$

(3)　$2x^2-8x+10$

(4)　$-4x^2+8x-3$

(5)　$5x^2+10x$

詳しい解説は下の方にあります。

LEVEL4

(1)　$3x^2+9x-1$

(2)　$2x^2-2x+9$

(3)　$4x^2+12x$

(4)　$5x^2+25x+28$

(5)　$6x^2+6x+1$

詳しい解説は下の方にあります。

LEVEL5

(1)　$-{\displaystyle \frac{1}{3}{x}^2+3x+1}$

(2)　${\displaystyle \frac{2}{5}{x}^2+4x-2}$

(3)　${\displaystyle \frac{1}{5}{x}^2+{\displaystyle \frac{2}{3}x}}$

(4)　$x^2+2ax+2$

(5)　$a{x}^2+2(a+1)x+2$

詳しい解説は下の方にあります。

解説

LEVEL1

(1)　$x^2-4x+7$

$=(x^2-4x+4)-4+7$
$=(x-2{)}^2+3$

(2)　$x^2+10x+19$

$=(x^2+10x+25)-25+19$
$=(x+5{)}^2-6$

(3)　$x^2-2x+3$

$=(x^2-2x+1)-1+3$
$=(x-1{)}^2+2$

(4)　$x^2-4x+4$

$=(x^2-4x+4)-4+4$
$=(x-2{)}^2$

※　この式の場合、平方完成した時と因数分解した時の式が同じになる。

(5)　$x^2+8x+11$

$=(x^2+8x+16)-16+11$
$=(x+4{)}^2-5$

(6)　$x^2-6x+7$

$=(x^2-6x+9)-9+7$
$=(x-3{)}^2-2$

(7)　$x^2-6x$

$=(x^2-6x+9)-9$
$=(x-3{)}^2-9$

(8)　$x^2+12x$

$=(x^2+12x+36)-36$
$=(x+6{)}^2-36$

↓平方完成の詳しい解説はこちら

LEVEL2

(1)　$x^2+3x+1$

$={(x+{\displaystyle \frac{3}{2}})}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2+1$
$={(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{9}{4}+1}$
$={(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{5}{4}}$

(2)　$x^2-7x$

$={(x+{\displaystyle \frac{7}{2}})}^2-{\left({\displaystyle \frac{7}{2}}\right)}^2$
$={(x-{\displaystyle \frac{7}{2}})}^2-{\displaystyle \frac{49}{4}}$

(3)　$x^2-5x+4$

$={(x+{\displaystyle \frac{5}{2}})}^2-{\left({\displaystyle \frac{5}{2}}\right)}^2+4$
$={(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{25}{4}+4}$
$={(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{9}{4}}$

(4)　$x^2+x-4$

$={(x+{\displaystyle \frac{1}{2}})}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2-4$
$={(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{4}+1}$
$={(x+{\displaystyle \frac{1}{2}})}^2+{\displaystyle \frac{3}{4}}$

(5)　$x^2-11x+8$

$={(x-{\displaystyle \frac{11}{2}})}^2-{\left({\displaystyle \frac{11}{2}}\right)}^2+8$
$={(x-{\displaystyle \frac{11}{2}})}^2-{\displaystyle \frac{121}{4}+8}$
$={(x-{\displaystyle \frac{11}{2}})}^2-{\displaystyle \frac{89}{4}}$

↓平方完成の詳しい解説はこちら

LEVEL3

(1)　$2x^2+4x-1$

$=2(x^2+2x)-1$
$=2\{(x^2+2x+1)-1\}-1$
$=2\{(x+1{)}^2-1\}-1$
$=2(x+1{)}^2-1\cdot 2-1$
$=2(x+1{)}^2-2-1$
$=2(x+1{)}^2-3$

(2)　$3x^2-12x+3$

$=3(x^2-4x)+3$
$=3\{(x^2-4x+4)-4\}+3$
$=3\{(x+2{)}^2-4\}+3$
$=3(x+2{)}^2-4\cdot 3+3$
$=3(x+2{)}^2-12+3$
$=3(x+2{)}^2-9$

(3)　$2x^2-8x+10$

$=2(x^2-4x)+10$
$=2\{(x^2-4x+4)-4\}+10$
$=2\{(x+2{)}^2-4\}+10$
$=2(x+2{)}^2-4\cdot 2+10$
$=2(x+2{)}^2-8+10$
$=2(x+2{)}^2+2$

(4)　$-4x^2+8x-3$

$=-4(x^2-2x)-3$
$=-4\{(x^2-2x+1)-1\}-3$
$=-4\{(x-1{)}^2-1\}-3$
$=-4(x-1{)}^2-1\cdot (-4)-3$
$=-4(x-1{)}^2+4-3$
$=-4(x-1{)}^2+1$

(5)　$5x^2+10x$

$=5(x^2+2x)$
$=5\{(x^2+2x+1)-1\}$
$=5\{(x+1{)}^2-1\}$
$=5(x+1{)}^2-1\cdot 5$
$=5(x+1{)}^2-5$

↓平方完成の詳しい解説はこちら

LEVEL4

(1)　$3x^2+9x-1$

$=3(x^2+3x)-1$
$=3\{(x^2+3x+{\displaystyle \frac{9}{4}})-{\displaystyle \frac{9}{4}}\}-1$
$=3\{{(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{9}{4}}\}-1$
$=3{(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{9}{4}\cdot 3-1}$
$=3{(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{27}{4}-1}$
$=3{(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{31}{4}}$

(2)　$2x^2-2x+9$

$=2(x^2-x)+9$
$=2\{(x^2-x+{\displaystyle \frac{1}{4}})-{\displaystyle \frac{1}{4}}\}+9$
$=2\{{(x-{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{4}}\}+9$
$=2{(x-{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{4}\cdot 2+9}$
$=2{(x-{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{2}+9}$
$=2{(x-{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{17}{2}}$

(3)　$4x^2+12x$

$=4(x^2+3x)$
$=4\{(x^2+3x+{\displaystyle \frac{9}{4}})-{\displaystyle \frac{9}{4}}\}$
$=4\{{(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{9}{4}}\}$
$=4{(x+{\displaystyle \frac{3}{2}})}^2-9$
$=4{(x+{\displaystyle \frac{3}{2}})}^2-9$

(4)　$5x^2+25x+28$

$=5(x^2+5x)+28$
$=5\{(x^2+5x+{\displaystyle \frac{25}{4}})-{\displaystyle \frac{25}{4}}\}+28$
$=5\{{(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{25}{4}}\}+28$
$=5{(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{25}{4}\cdot 5+28}$
$=5{(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{125}{4}+28}$
$=5{(x+{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{13}{4}}$

(5)　$6x^2+6x+1$

$=6(x^2+x)+1$
$=6\{(x^2+x+{\displaystyle \frac{1}{4}})-{\displaystyle \frac{1}{4}}\}+1$
$=6\{{(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{4}}\}+1$
$=6{(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{4}\cdot 6+1}$
$=6{(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{3}{2}+1}$
$=6{(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{1}{2}}$

↓平方完成の詳しい解説はこちら

LEVEL5

(1)　$-{\displaystyle \frac{1}{3}{x}^2+3x+1}$

$=-{\displaystyle \frac{1}{3}(x^2-9x)+1}$
$=-{\displaystyle \frac{1}{3}(x^2-9x+{\displaystyle \frac{81}{4}})-{\displaystyle \frac{81}{4}+1}}$
$=-{\displaystyle \frac{1}{3}\{{(x-{\displaystyle \frac{9}{2}})}^2-{\displaystyle \frac{81}{4}}\}+1}$
$=-{\displaystyle \frac{1}{3}{(x+{\displaystyle \frac{9}{2}})}^2-{\displaystyle \frac{81}{4}\cdot (-\frac{1}{3})+1}}$
$=-{\displaystyle \frac{1}{3}{(x+{\displaystyle \frac{9}{2}})}^2+{\displaystyle \frac{27}{4}+1}}$
$=-{\displaystyle \frac{1}{3}{(x+{\displaystyle \frac{9}{2}})}^2+{\displaystyle \frac{31}{4}}}$

(2)　${\displaystyle \frac{2}{5}{x}^2+4x-2}$

$={\displaystyle \frac{2}{5}(x^2+10x)-2}$
$={\displaystyle \frac{2}{5}(x^2+10x+25)-25-2}$
$={\displaystyle \frac{2}{5}\{(x+5{)}^2-25\}-2}$
$={\displaystyle \frac{2}{5}(x+5{)}^2-25\cdot {\displaystyle \frac{2}{5}-2}}$
$={\displaystyle \frac{2}{5}(x+5{)}^2-10-2}$
$={\displaystyle \frac{2}{5}(x+5{)}^2-12}$

(3)　${\displaystyle \frac{1}{5}{x}^2+{\displaystyle \frac{2}{3}x}}$

$={\displaystyle \frac{1}{5}(x^2+{\displaystyle \frac{10}{3}x})}$
$={\displaystyle \frac{1}{5}(x^2+{\displaystyle \frac{10}{3}x+\frac{25}{9}})-\frac{25}{9}}$
$={\displaystyle \frac{1}{5}\{{(x+{\displaystyle \frac{5}{3}})}^2-{\displaystyle \frac{25}{9}}\}}$
$={\displaystyle \frac{1}{5}{(x+{\displaystyle \frac{5}{3}})}^2-{\displaystyle \frac{25}{9}\cdot \frac{1}{5}}}$
$={\displaystyle \frac{1}{5}{(x+{\displaystyle \frac{5}{3}})}^2-{\displaystyle \frac{5}{9}}}$

(4)　$x^2+2ax+2$

$=(x+2ax+a^2)-a^2+2$
$=(x+a{)}^2-a^2+2$

(5)　$a{x}^2+2(a+1)x+2$

$=a(x^2+{\displaystyle \frac{2(a+1)}{a}x})+2$
$=a(x^2+{\displaystyle \frac{2(a+1)}{a}x+{\left(\frac{a+1}{a}\right)}^2})-{\left({\displaystyle \frac{a+1}{a}}\right)}^2+2$
$=a\{{(x+{\displaystyle \frac{a+1}{a}})}^2-{\left({\displaystyle \frac{a+1}{a}}\right)}^2\}+2$
$=a{(x+{\displaystyle \frac{a+1}{a}})}^2-{\left({\displaystyle \frac{a+1}{a}}\right)}^2\cdot a+2$
$=a{(x+{\displaystyle \frac{a+1}{a}})}^2-{\displaystyle \frac{(a+1{)}^2}{a}+2}$

↓平方完成の詳しい解説はこちら