Question

Frame Quadratic equations using the following roots & then verify.find the answers with process please (class 10)
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Answer 1

Solution:

[i] (-5/7,2)

Roots are -5/7  and 2

Sum of Roots , $S = \frac{-5}{7} + 2 = \frac{-5 + 14}{7} = \frac{9}{7}$

Product of Roots, $P = (\frac{-5}{7}) * (2) = \frac{-10}{7}$

Required equation = $x^{2} - Sx + P$ = $x^{2} - \frac{9}{7} x + (\frac{-10}{7})$ = $x^{2} - \frac{9}{7} x - \frac{10}{7}$

[ii] (3,-2/3)

Roots are 3 and -2/3

Sum of Roots , $S = 3 + \frac{-2}{3} = \frac{9 - 2}{3} = \frac{7}{3}$

Product of Roots, $P = (3) * (\frac{-2}{3}) = \frac{-6}{3} = -2$

Required equation = $x^{2} - Sx + P$ = $x^{2} - \frac{7}{3} x + (-2)$ = $x^{2} - \frac{7}{3} x - 2$

[iii] (-1/5,1/2)

Roots are -1/5 and 1/2

Sum of Roots , $S = \frac{-1}{5} + \frac{1}{2} = \frac{-2 + 5}{10} = \frac{3}{10}$

Product of Roots, $P = (\frac{-1}{5}) * (\frac{1}{2}) = \frac{-1}{10}$

Required equation = $x^{2} - Sx + P$ = $x^{2} - \frac{3}{10} x + (\frac{-1}{10})$ = $x^{2} - \frac{3}{10} x - \frac{1}{10}$

[iv] (-2/5,-1/2)

Roots are -2/5 and -1/2

Sum of Roots , $S = \frac{-2}{5} + \frac{-1}{2} = \frac{-4 -5}{10} = \frac{-9}{10}$

Product of Roots, $P = (\frac{-2}{5}) * (\frac{-1}{2}) = \frac{2}{10} = \frac{1}{5}$

Required equation = $x^{2} - Sx + P$ = $x^{2} - (\frac{-9}{10}) x + \frac{1}{5}$ = $x^{2} + \frac{9}{10} x + \frac{1}{5}$

Explanation:

Let α and β be the two roots to form a quadratic equation

Let required equation be ax² + bx + c = 0 (a ≠ 0).

Roots of this equation are α and β.  

Therefore,

α + β = -b/a and αβ = c/a.

Now,

ax² + bx + c = 0

x² - (α + β)x + αβ = 0, [Since, α + β = -b/a and αβ = c/a]

x² - (Sum of roots)x + Product of roots = 0

x² - Sx + P = 0   [Used for forming quadratic equations]

where,

S = Sum of the roots

P = Product of the roots