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Quadratic Formula

Find the root...

Question

Find the roots of the quadratic equations by using the quadratic formula in the following equation:
(i) $2 x^{2} - 3 x - 5 = 0$
(ii) $5 x^{2} + 13 x + 8 = 0$
(iii) $- 3 x^{2} + 5 x + 12 = 0$
(iv) $- x^{2} + 7 x - 10 = 0$
(v) $x^{2} + 2 \sqrt{2} x - 6 = 0$
(vi) $x^{2} - 3 \sqrt{5} x + 10 = 0$
(vii) $\frac{1}{2} x^{2} - \sqrt{11} x + 1 = 0$

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Solution

(i) Given equation is $2 x^{2} - 3 x - 5 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = 2, b = - 3 and c = - 5$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$
$= \frac{- (- 3) \pm \sqrt{(- 3)^{2} - 4 (2) (- 5)}}{2 (2)}$
$= \frac{3 \pm \sqrt{9 + 40}}{4}$
$= \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}$
$\Rightarrow x = \frac{3 + 7}{4} o r x = \frac{3 - 7}{4}$
$\Rightarrow x = \frac{10}{4} o r x = - \frac{4}{4}$
$\Rightarrow x = \frac{5}{2} o r x = - 1$
Hence, the roots of the given equation are $\frac{5}{2} and - 1$ .

(ii) Given equation is $5 x^{2} + 13 x + 8 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = 5, b = 13 and c = 8$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ ,
$= \frac{- (13) \pm \sqrt{(13)^{2} - 4 (5) (8)}}{2 (5)}$
$= \frac{- 13 \pm \sqrt{169 - 160}}{10}$
$= \frac{- 13 \pm \sqrt{9}}{10} = \frac{- 13 \pm 3}{10}$
$\Rightarrow x = \frac{- 13 + 3}{10} o r x = \frac{- 13 - 3}{10}$
$\Rightarrow x = - \frac{10}{10} o r x = \frac{- 16}{10}$
$\Rightarrow x = - 1 o r x = - \frac{8}{5}$
Hence, the roots of the given equation are $- 1 and \frac{- 8}{5}$ .

(iii) Given equation is $- 3 x^{2} + 5 x + 12 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = - 3, b = 5 and c = 12$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ ,
$= \frac{- 5 \pm \sqrt{(5)^{2} - 4 (- 3) (12)}}{2 (- 3)}$
$= \frac{- 5 \pm \sqrt{25 + 144}}{- 6} = \frac{- 5 \pm \sqrt{169}}{- 6}$
$= \frac{- 5 \pm 13}{- 6}$
$\Rightarrow x = (\frac{- 5 + 13}{- 6}) o r x = (\frac{- 5 - 13}{- 6})$
$\Rightarrow x = \frac{8}{- 6} o r x = \frac{- 18}{- 6}$
$\Rightarrow x = - \frac{4}{3} o r x = 3$
Hence, the roots of the given equation are $- \frac{4}{3} and 3.$

(iv) Given equation is $- x^{2} + 7 x - 10 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = - 1, b = 7 and c = - 10$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ ,
$= \frac{- 7 \pm \sqrt{(7)^{2} - 4 (- 1) (- 10)}}{2 (- 1)}$
$= \frac{- 7 \pm \sqrt{49 - 40}}{- 2} = \frac{- 7 \pm \sqrt{9}}{- 2}$
$= \frac{- 7 \pm 3}{- 2}$
$\Rightarrow x = \frac{- 7 + 3}{- 2} o r x = \frac{- 7 - 3}{- 2}$
$\Rightarrow x = \frac{- 4}{- 2} o r x = \frac{(- 10)}{(- 2)}$
$\Rightarrow x = 2 o r x = 5$
Hence, the roots of the given equation are 2 and 5.

(v) Given equation is $x^{2} + 2 \sqrt{2} x - 6 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = 1, b = 2 \sqrt{2} and c = - 6$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ ,
Therefore, roots of given quadratic equation is given by
$x = \frac{- (2 \sqrt{2}) \pm \sqrt{(2 \sqrt{2})^{2} - 4 (1) (- 6)}}{2 (1)}$
$= \frac{- 2 \sqrt{2} \pm \sqrt{8 + 24}}{2}$
$= \frac{- 2 \sqrt{2} \pm \sqrt{16 \times 2}}{2} = \frac{- 2 \sqrt{2} \pm 4 \sqrt{2}}{2}$
$= - \sqrt{2} \pm 2 \sqrt{2}$
$\Rightarrow x = - \sqrt{2} + 2 \sqrt{2} o r x = - \sqrt{2} - 2 \sqrt{2}$
$\Rightarrow x = \sqrt{2} o r x = - 3 \sqrt{2}$
Hence, the roots of the given equation are $\sqrt{2} and - 3 \sqrt{2}$

(vi) Given equation is $x^{3} - 3 \sqrt{5} x + 10 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = 1, b = - 3 \sqrt{5} and c = 10$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$
Therefore, roots of given quadratic equation is given by
$x = \frac{- (- 3 \sqrt{5}) \pm \sqrt{(- 3 \sqrt{5})^{2} - 4 (1) (10)}}{2 (10)}$
$= \frac{3 \sqrt{5} \pm \sqrt{45 - 40}}{2} = \frac{3 \sqrt{5} \pm \sqrt{5}}{2}$
$\Rightarrow x = \frac{3 \sqrt{5} + \sqrt{5}}{2} o r x = \frac{3 \sqrt{5} - \sqrt{5}}{2}$
$\Rightarrow x = 2 \sqrt{5} o r x = \sqrt{5}$
Hence, the roots of the given equation are $2 \sqrt{5} and \sqrt{5}$ .

(vii) Given equation is $\frac{1}{2} x^{2} - \sqrt{11} x + 1 = 0$
On comparing with $a x^{2} + b x + c = 0,$ we get $a = \frac{1}{2}, b = - \sqrt{11} and c = 1$
Quadratic formula, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$
Therefore, roots of given quadratic equation is given by
$x = \frac{- (- \sqrt{11}) \pm \sqrt{(- \sqrt{11})^{2} - 4 \times \frac{1}{2} \times 1}}{2 (\frac{1}{2})}$
$= \frac{\sqrt{11} \pm \sqrt{11 - 2}}{(1)}$
$= \sqrt{11} \pm \sqrt{9}$
$\Rightarrow x = \sqrt{11} + 3 o r x = \sqrt{11} - 3$
Hence, the roots of the given equation are
$\sqrt{11} + 3 and \sqrt{11} - 3$ .

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