1sec ≫ First term ≫ Algebra ≫ Unit 1 ≫(1 – 4 ) The Relation Between the Roots of the Second Degree Equation and the Coefficients of its Terms.
(Q1 to Q10)
(1) Without solving the equation, find the sum and the product of the two roots of the equation \(2x^2 + 5x - 12 = 0\).
Solution on YouTubeThe correct answer is: Sum of the roots = \(-\frac{5}{2}\) and Product of the roots = \(-6\).
(2) If the product of the two roots of the equation \(2x^2 - 3x + k = 0\) equals 1, find the value of \(k\), then solve the equation.
The product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula:
\[ \text{Product of roots} = \frac{c}{a} \]For the equation \(2x^2 - 3x + k = 0\), \(a = 2\), \(b = -3\), and \(c = k\). So, according to the given condition:
\[ \frac{k}{2} = 1 \] \[ k = 2 \]Now, to solve the equation \(2x^2 - 3x + 2 = 0\), we can use the quadratic formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]Substituting the values \(a = 2\), \(b = -3\), and \(c = 2\), we get:
\[ x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 2 \cdot 2}}}}{{2 \cdot 2}} \] \[ x = \frac{{3 \pm \sqrt{9 - 16}}}{{4}} \] \[ x = \frac{{3 \pm \sqrt{-7}}}{{4}} \]Since the discriminant (\(b^2 - 4ac\)) is negative, the roots are complex numbers.
(3) In the quadratic equation: $b x^{2} + c x + a = 0$, if the sum of the two roots equals the product of them, then $\mathrm{c} = $
- (a) b
- (b) a
- (c) $-\mathrm{b}$
- (d) $-\mathrm{a}$
The correct answer is (d) $-\mathrm{a}$
Solution on YouTube(4) If $\mathrm{L}, \mathrm{M}$ are the two roots of the equation : $x^{2}+x+1=0$, then $\mathrm{L}+\mathrm{M}+\mathrm{LM}=$
- (a) zero
- (b) 1
- (c) -1
- (d) 2
The correct answer is (a) zero
Solution on YouTube(5) If one of the two roots of the equation: $x^{2}-(b-3) x+5=0$ is the additive inverse of the other root, then $\mathrm{b} =$
- (a) -5
- (b) -3
- (c) 3
- (d) 5
The correct answer is (c) 3
Solution on YouTube(6) If one of the two roots of the equation : $ax^{2}-3x+2=0$ is the multiplicative inverse of the other, then $\mathrm{a} =$
- (a) $\frac{1}{3}$
- (b) $\frac{1}{2}$
- (c) 2
- (d) 3
The correct answer is (a) $\frac{1}{3}$
Solution on YouTube(7) Without solving the equation, find the sum and the product of the two roots of the following equations : $3x^{2}=23x-30$
The correct answer is $sum = \frac{23}{3}$ and $product = \frac{10}{1}$
Solution on YouTube(8) Without solving the equation, find the sum and the product of the two roots of the following equation:
$\frac{x}{2}+\frac{1}{x}=\frac{3}{2}$
The correct answer is $sum= 3$ and $product = 2$
Solution on YouTube(9) If the product of the two roots of the equation: $3x^{2} + 10x - c = 0$ is $\frac{-8}{3}$, find the value of $c$, then solve the equation in the set of complex numbers.
The correct answer is: $$ \mathrm{c} = 8, \, x = \frac{2}{3} \text{ or } x = -4 $$
Solution on YouTube(10) If the sum of the two roots of the equation: \(2 x^{2} + b x - 5 = 0\) is \(\frac{-3}{2}\), find the value of \(b\), then solve the equation in the set of complex numbers.
The correct answer is:
\[ b = 3, x = \frac{-5}{2} \text{ or } x = 1 \]
No comments:
Post a Comment