QUADRATIC FUNCTIONS : EXERCISE (5.4) SOLUTIONS



            Solve the following equations by using the quadratic formula.

1.          $x^{2}+2 x-1=0$

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$\begin{array}{l} \ \ \ \ {{x}^{2}}+2x-1=0\\\\ \ \ \ \ \text{Comparing with}\ a{{x}^{2}}+bx+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=2,\ c=-1\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{2}^{2}}-4(2)(-1)=8>0\\\\ \ \ \ \ \therefore \ \text{There are two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm \sqrt{{{{2}^{2}}-4(2)(-1)}}}}{{2(1)}}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm \sqrt{8}}}{2}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm 2\sqrt{2}}}{2}\\\\ \ \ \ \ \ \ \ \ \ =-1\pm \sqrt{2}\\\\ \ \ \ \ \therefore \ \ x=-1-\sqrt{2}\ \text{or}\ x=-1+\sqrt{2}\end{array}$

2.          $x^{2}+4 x+4=0$

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$ \begin{array}{l}\ \ \ {{x}^{2}}+4x+4=0\\\\\;\;\;\;\text{Comparing with}\;a{{x}^{2}}+bx+c=0,\;\text{we get}\;\\\;\;\;\\\ \ \ a=1,\;b=4,\;c=4\;\;\\\;\;\\\ \ \ \text{discriminant}={{b}^{2}}-4ac={{4}^{2}}-4(1)(4)=0\;\\\;\;\;\\\ \ \therefore \;\text{There is one real solution for }x.\;\;\;\\\;\\\ \ \therefore \;\;x=-\displaystyle \frac{b}{{2a}}=-\displaystyle \frac{4}{{2(1)}}=-2\end{array}$

3.          $p^{2}-6 p+3=0$

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$ \begin{array}{l}\ \ \ \ {{p}^{2}}-6p+3=0\\\\ \ \ \ \ \text{Comparing with}\ a{{p}^{2}}+bp+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=-6,\ c=3\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-6)}^{2}}-4(1)(3)=24>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{6\pm \sqrt{{24}}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{6\pm 2\sqrt{6}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =3\pm \sqrt{6}\\\\ \ \ \ \ \therefore \ \ x=3-\sqrt{6}\ \text{or}\ x=3+\sqrt{6}\end{array}$

4.          $t^{2}-4 t-8=0$

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$\begin{array}{l} \ \ \ \ {{t}^{2}}-4t-8=0\\\\ \ \ \ \ \text{Comparing with}\ a{{t}^{2}}+bt+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=-4,\ c=-8\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-4)}^{2}}-4(1)(-8)=48>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4\pm \sqrt{{48}}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4\pm 4\sqrt{3}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =2\pm 2\sqrt{3}\\\\ \ \ \ \ \therefore \ \ x=2-2\sqrt{3}\ \text{or}\ x=2+2\sqrt{3} \end{array}$

5.          $3 q^{2}-12 q+11=0$

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$\begin{array}{l}\ \ \ \ 3{{q}^{2}}-12q+11=0\\\\ \ \ \ \ \text{Comparing with}\ a{{q}^{2}}+bq+c=0,\ \text{we get}\\\\ \ \ \ \ a=3,\ b=-12,\ c=11\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-12)}^{2}}-4(3)(11)=12>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{12\pm \sqrt{{12}}}}{6}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{12\pm 3\sqrt{3}}}{6}\\\\ \ \ \ \ \ \ \ \ \ \ =2\pm \displaystyle \frac{{\sqrt{3}}}{2}\\\\ \ \ \ \ \therefore \ \ x=2-\displaystyle \frac{{\sqrt{3}}}{2}\ \text{or}\ x=2+\displaystyle \frac{{\sqrt{3}}}{2} \end{array}$

6.          $5 z^{2}+3 z-4=0$

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$\begin{array}{l} \ \ \ \ 5{{z}^{2}}+3z-4=0\\\\ \ \ \ \ \text{Comparing with}\ a{{z}^{2}}+bz+c=0,\ \text{we get}\\\\ \ \ \ \ a=5,\ b=3,\ c=-4\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{3}^{2}}-4(5)(-4)=89>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-3\pm \sqrt{{89}}}}{{10}}\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-3-\sqrt{{89}}}}{{10}}\ \text{or}\ x=\displaystyle \frac{{-3+\sqrt{{89}}}}{{10}} \end{array}$