【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)　\(ax^2+2(a+1)x+2\)

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)　\(x^2+3x+1\)

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

(2)　\(x^2-7x\)

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

(3)　\(x^2-5x+4\)

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

(4)　\(x^2+x-4\)

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

(5)　\(x^2-11x+8\)

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

↓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)　\(3x^2+9x-1\)

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

(2)　\(2x^2-2x+9\)

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

(3)　\(4x^2+12x\)

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

(4)　\(5x^2+25x+28\)

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

(5)　\(6x^2+6x+1\)

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

↓平方完成の詳しい解説はこちら

LEVEL5

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

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

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

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

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

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

(4)　\(x^2+2ax+2\)

\(=(x+2ax+a^2)-a^2+2\)
\(=(x+a)^2-a^2+2\)

(5)　\(ax^2+2(a+1)x+2\)

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

↓平方完成の詳しい解説はこちら