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  • Find the roots of the quadratic equations by using the quadratic formula in each of the following: (i) 2x square minus 3x minus 5 is equal to 0 (ii)5x square plus 13x plus 8 equal to 0 (iii)minus 3x square plus 5x plus 12 equal to 0

Find the roots of the quadratic equations by using the quadratic formula in each of the following:

(i)2x^{2}-3x-5=0

(ii)5x^{2}+13x+8=0       

(iii)-3x^{2}+5x+12=0

(iv)-x^{2}+7x-10=0

(v)x^{2}+2\sqrt{2}x-6=0

(vi)x^{2}+3\sqrt{5}x+10=0

(vii)\frac{1}{2}x^{2}-\sqrt{11}x+1=0

Answers (1)

(i) \frac{5}{2},-1

2x^{2}-3x-5=0

Compare with ax^{2}+bx+c=0 where a \neq 0

Here    a=2, b=-3, c=-5
 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-(-3)\pm \sqrt{(-3)^{2}-4(2)(-5)}}{2 \times 2}\\ x=\frac{3 \pm \sqrt{9+40}}{4}\\ x=\frac{3 \pm 7}{4}\\ x=\frac{3+7}{4}=\frac{10}{4}=\frac{5}{2 }\;\; \; \;\;\; x=\frac{3-7}{4} =\frac{-4}{4} =-1

 x=\frac{5}{2},-1

 \left (\frac{5}{2},-1 \right )  are the roots of the equation 2x^{2}-3x-5=0

 

(ii) -1,\frac{-8}{5}

 5x^{2}+13x+8=0

 Compare with ax^{2}+bx+c=0 where a \neq 0

 Here    a=5, b=13, c=8
 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-13\pm \sqrt{(13)^{2}-4(5)(8)}}{2 \times 5}\\ x=\frac{-13 \pm \sqrt{169-160}}{10}\\ x=\frac{-13 \pm 3}{10}\\ x=\frac{-13+3}{10}=\frac{-10}{10}=-1\;\; \; \;\;\; x=\frac{-13-3}{10} =\frac{-16}{10} =\frac{-8}{5}

                   x=-1,\frac{-8}{5}

 \left (-1,\frac{-8}{5} \right )  are the roots of the equation 5x^{2}+13x+8=0

(iii) \frac{-4}{3},3

-3x^{2}+5x+12=0

Compare with ax^{2}+bx+c=0 where a \neq 0

Here    a=-3, b=5, c=12

 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-5\pm \sqrt{(5)^{2}-4(-3)(12)}}{2 \times (-3)}\\ x=\frac{-5 \pm \sqrt{25+144}}{-6}\\ x=\frac{-5 \pm 13}{-6}\\ x=\frac{-5+13}{-6}=\frac{8}{-6}=\frac{-4}{3 }\;\; \; \;\;\; x=\frac{-5-13}{-6} =\frac{-18}{-6} =3

 x=\frac{-4}{3},3

 \left (\frac{-4}{3},3 \right )  \text{are the roots of the equation} -3x^{2}+5x+12=0

(iv) 5,2

 -x^{2}+7x-10=0

Compare with ax^{2}+bx+c=0 where a \neq 0

Here    a=-1, b=7, c=-10

 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-7\pm \sqrt{49-4(-1)(-10)}}{2 \times (-1)}\\ x=\frac{-7 \pm \sqrt{9}}{-2}\\ x=\frac{-7 \pm 3}{-2}\\ x=\frac{-7+3}{-2}=\frac{-4}{-2}=2\;\; \; \;\;\; x=\frac{-7-3}{-2} =\frac{-10}{-2} =5

 x=5,2

 (5,2)  are the roots of the equation -x^{2}+7x-10=0

(v) \sqrt{2},-3\sqrt{2}

x^{2}+2\sqrt{2}x-6=0

Compare with ax^{2}+bx+c=0 where a \neq 0           

Here    a=1, b=2\sqrt{2}, c=-6

 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-2\sqrt{2}\pm \sqrt{(2\sqrt{2})^{2}-4(1)(-6)}}{2 \times 1}\\ x=\frac{-2\sqrt{2} \pm \sqrt{8+24}}{2}\\ x=\frac{2\sqrt{2} \pm 4\sqrt{2}}{2}\\ x=\frac{2\left (\sqrt{2} \pm 2 \sqrt{2} \right )}{2}\\ x=-\sqrt{2}\pm2\sqrt{2}\\ x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\;\; \; \;\;\; x=-\sqrt{2}-2\sqrt{2}=-3\sqrt{2}

                   x=\sqrt{2},-3\sqrt{2}

            \sqrt{2},-3\sqrt{2}  are the roots of the equation x^{2}+2\sqrt{2}x-6=0

(vi) 2\sqrt{5},\sqrt{5}

x^{2}+3\sqrt{5}x+10=0

Compare with ax^{2}+bx+c=0 where a \neq 0           

a=1, b=3\sqrt{5}, c=10

 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{3\sqrt{5}\pm \sqrt{(3\sqrt{5})^{2}-4(1)(10)}}{2 \times 1}\\ x=\frac{3\sqrt{5} \pm \sqrt{45-40}}{2}\\ x=\frac{3\sqrt{5} \pm \sqrt{5}}{2}\\ x=\frac{3\sqrt{5} + \sqrt{5}}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}\;\; \; \;\;\; x=\frac{3\sqrt{5} - \sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}

 x=2\sqrt{5},\sqrt{5}

 2\sqrt{5},\sqrt{5}  are the roots of the equation x^{2}+3\sqrt{5}x+10=0

(vii) \sqrt{11}+3,\sqrt{11}-3

 \frac{1}{2}x^{2}-\sqrt{11}x+1=0

Compare with ax^{2}+bx+c=0 where a \neq 0           

 Here a=\frac{1}{2},b=-\sqrt{11},c=1

 \\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{\sqrt{11}\pm \sqrt{(-\sqrt{11})^{2}-4\left ( \frac{1}{2} \right )(1)}}{2 \times \frac{1}{2}}\\ x=\frac{\sqrt{11}\pm\sqrt{11-2}}{1}\\ x=\sqrt{11}\pm 3\\ x=\sqrt{11}+ 3\;\;\;\;\;\;\;x=\sqrt{11}- 3

 x=\left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right )

hence \left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right ) are the root of the equation \frac{1}{2}x^{2}-\sqrt{11}x+1=0

Posted by

infoexpert24

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