Question

Find the zeroes of the polynomial x2 + 3x – 2, and verify the
relation between the coefficients and the zeroes of the polynomial.
​

Answer 1

Answer:

$x = - 1 \: or \: - 2$

Step-by-step explanation:

${x}^{2} + 3x - 2 \\ {x}^{2} + x + 2x - 2 \\ x(x + 1) + 2(x + 1) \\ (x + 2)(x + 1) \\ equating \: each \: to \: zero \\ x + 2 = 0$

$x = - 2 \\ or \\ x + 1 = 0 \\ x = - 1$

$veryfying \: \: \: relationship$

$a = 1 \\ b = 3 \\ c = - 2$

$let \: \alpha \: = - 1 \: and \: \beta = - 2 \\ \alpha + \beta = \frac{ - b}{a} \\ \alpha + \beta = - 1 + ( - 2) \\ = - 1 - 2 \\ = - 3$

$\frac{ - b}{a} \\ \frac{ - 3}{1} \\ = - 3 \\ so \: \: l.h.s \: \: = \: \: r.h.s$

$\alpha \beta = \frac{c}{a} \\ \alpha \beta = ( - 1)( - 2) \\ = - 2 \\ \frac{c}{a} = \frac{ - 2}{1} \\ = - 2 \\ so \: l.h.s \: \: = \: \: r.h.s$

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Answer 2

Solution :-

${x}^{2} + 3x - 2$

We can't use mid term factoring theorem in this Polynomial so we'll have to use the quadratic formula .

$x = \frac{ - b + \sqrt{ {b }^{2} - 4ac } }{2a}$

Here, By comparing the given polynomial with ax² + bx +c

$\star \: a = 1 \\ \\ \star \: b \: = 3 \\ \\ \star \: c = - 2$

$: \implies \: x = \frac{ - 3 + \sqrt{( {3)}^{2} - 4 \times 1 \times ( - 2) } }{2 \times 1} \\ \\ : \implies \: x = \frac{ - 3 + \sqrt{9 + 8} }{2} \\ \\ : \implies \: x = \frac{ - 3 + \sqrt{17} }{2}$

So, the zeros are →

$\star \: x = \frac{ - 3 + \sqrt{17} }{2} \\ \\ \star \: x = \frac{ - 3 - \sqrt{17} }{2}$

The relationship between coefficients and the zeroes :-

Let the zeros are α and β , so

$\star \: \alpha = \frac{ - 3 + \sqrt{17} }{2} \\ \\ \star \: \beta = \frac{ - 3 - \sqrt{17} }{2}$

• First relation →

$\alpha + \beta = - \frac{b}{a}$

$\implies \frac{ - 3 + \sqrt{17} }{2} + \frac{ - 3 - \sqrt{17} }{2} = - \frac{3}{1} \\ \\ \implies \: \frac{ - 3 + \sqrt{17} - 3 - \sqrt{17} }{2} = - 3 \\ \\ \implies \: \frac{ - 6}{2} = - 3 \\ \\ \implies \: - 3 = - 3 \\ \\ l.h.s. = r.h.s$

• Second Relation →

$\alpha \beta = \frac{c}{a}$

$\implies \frac{ - 3 + \sqrt{17} }{2} \times \frac{ - 3 - \sqrt{17} }{2} = \frac{ - 2}{1} \\ \\ \implies \: \frac{( -3 + \sqrt{17} )( - 3 - \sqrt{17} )}{4} = - 2 \\ \\ \implies \: \frac{( \sqrt{17} - 3) \times - ( \sqrt{17} + 3) }{4} = - 2 \\ \\ \implies \: \frac{ - ( \sqrt{17} - 3)( \sqrt{17} + 3) }{4} = - 2 \\ \\ \implies \: - \frac{( { \sqrt{17}) }^{2} - ( {3)}^{2} }{4} = - 2 \\ \\ \implies \: - \frac{17 - 9}{4} = - 2 \\ \\ \implies \: - \frac{8}{4} = - 2 \\ \\ \implies \: - 2 = - 2 \\ \\ l.h.s. = r.h.s.$

Hence Proved !!