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August 11, 2023
When it comes to swiftly solving quadratic equations, a few key tips can significantly expedite the process. Start by recognizing opportunities for efficient factoring, which can unveil the equation’s roots effortlessly. Should factoring pose a challenge, leverage the quadratic formula—a potent tool providing precise solutions.
Solving quadratic equations quickly involves employing various methods that streamline the process and provide efficient solutions. Here are some effective methods:
Factoring: If possible, factor the quadratic equation into two binomials and set each binomial equal to zero. This method is particularly efficient when the equation is easily factorable.
Quadratic Formula: Utilize the quadratic formula to directly find the solutions of the equation. The formula is x = (-b ± √(b² – 4ac)) / 2a, where “a,” “b,” and “c” are the coefficients of the quadratic equation. This method is reliable and applicable to all quadratic equations.
Completing the Square: Transform the quadratic equation into a perfect square trinomial by adding and subtracting a suitable constant. This method is useful for equations that are not easily factorable, and it helps simplify the process of finding the roots.
Graphical Analysis: Graph the quadratic equation on a coordinate plane and determine the points where the graph intersects the x-axis. This method provides an approximate visual representation of the roots.
Use Symmetry: If one root of the quadratic equation is known, exploit the symmetry of the parabola to quickly deduce the other root.
Shortcut Techniques: For specific types of quadratic equations, such as those with special patterns (perfect squares, difference of squares), you can use shortcut techniques to simplify the solving process.
Mental Math: Develop mental math skills to perform calculations quickly and accurately. Simplify coefficients and terms mentally to expedite the calculations.
Where, x is the unknown variable and a, b, c are the numerical coefficients.
Example : Solve for x : x2-3x-10 = 0Solution : Let us express -3x as a sum of -5x and +2x.
x2-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
x-5 = 0 or x+2 = 0
x = 5 or x = -2
Question 1 Solve for the equations 17x2 + 48x – 9 = 0 and 13y2 – 32y + 12 = 0
Options:
A. x < y
B. x > y
C. x ≤ y
D. x ≥ y
E. cannot be determined
Solution 17x2 + 48x – 9 = 0…. (1)
17x2 + 51x – 3x – 9 = 0
(x+3) (17x – 3) = 0
Therefore, Roots of first equation are -3 and \frac{3}{17}
We know that, if sign given in the equation is + and – then their sign of roots is – and + respectively.
Therefore, the roots of the equation are – 3 and – \frac{3}{17}
Now, 13y2 – 32y + 12 = 0 ……. (2)
13y2 – 26y – 6y + 12 = 0
(y-2) (13y – 6) = 0
Therefore, Roots of second equation are 2 and \frac{6}{3}
We know that, If sign given in the equation is – and – then their sign of roots is + and + respectively.
Therefore, the roots of the equation are +2 and + \frac{6}{13}
Now, compare the roots – x1, + x2, + y1, and + y2
It means y>x
Correct option: A
Question 2. Solve for the equations \mathbf{\sqrt{500}x =\sqrt{420}} and \mathbf{\sqrt{260}y – \sqrt{200} = 0 }
Options
Solution: \sqrt{500}x -\sqrt{420} = 0 …… (1)
\sqrt{500}x =\sqrt{420}
500 x = 420
x =\frac{420}{500}
x = \frac{210}{250}
x = 0.84
Now, \sqrt{260}y -\sqrt{200} = 0
\sqrt{260}y = \sqrt{200}
260y = 200
y = \frac{200}{260}
y = \frac{100}{130}
y = \frac{5}{9}
y = 0.76
On comparing x and y, it is clear that x > y
Correct option: B
Question 3. Solve for the equations x ^{2} – 11x + 24 = 0 and 2y^{2}– 9y + 9 = 0
Solution: x^{2} – 11x + 24 = 0 ……….. (1)
x^{2} – 8x-3x + 24 = 0
(x – 8) (x – 3) = 0
Therefore, Roots of first equation are 8 and 3
We know that, If sign given in the equation is – and – then their sign of roots is + and +
Therefore, the roots of the equation are +8 and + 3
Now, 2y^{2}– 9y + 9 = 0 ……. (2)
2y^{2}– 6y – 3y+ 9 = 0
(y-3) (2y – 3) = 0
Therefore, Roots of second equation are 3 and 1.5
We know that, If sign given in the equation is – and + then their sign of roots is + and +
Therefore, the roots of the equation are + 3 and + 1.5
Now, compare the roots +x1, +x2, + y1, and + y2
It means x ≥ y
Correct option: D
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Question 4. Solve for equations 9x2 – 36x + 35 = 0 and 2y2 – 15y – 17 = 0
A. x < y B. x > y
Solution: 9x2 – 36x + 35 = 0……….. (1)
9x2 – 21x – 15x + 35 = 0
(3x – 7) (3x – 5) = 0
Therefore, Roots of first equation are \frac{7}{3} and \frac{5}{3}
We know that, If sign given in the equation is – and – then their sign of roots is + and + respectively
Therefore, the roots of the equation are +1.66 and +2.33
Now,
2y2 – 15y – 17 = 0……. (2)
2y2 – 17y + 2y – 17 = 0
(y + 1) (2y – 17) = 0
Therefore, Roots of second equation are 8.5 and -1
Now, compare the roots +x1, + x2, + y1, and – y2
It means , we cannot find any relation between x and y
Correct option: E
Question 5. Solve for equations x^{2} – 165 = 319 and y^{2} + 49 = 312
Solution x2– 165 = 319……(1)
x2– 165 – 319 = 0
x2= 484 = ±22
y2 + 49 = 312…. (2)
y2 + 49 -312 = 0
y2 = 263 y = ± 16.21
Now compare, x and y It means, we cannot find any relation between x and y
Question 3. Solve for Equations 4x2 + 18x – 10 = 0 and y^\frac{2}{5} – (\frac{25}{y})^\frac{8}{5} = 0
Solution: 4x2 + 18x – 10 = 0……….. (1)
Simplify it, 2x2 + 9x – 5 = 0
2x2 + 10x – x – 5 = 0
(x + 5) (2x – 1) = 0
Therefore, Roots of first equation are 5 and 0.5
We know that, if sign given in the equation is + and – then their sign of roots is – and +
Therefore, the roots of the equation are – 5 and +0.5
Now, y^\frac{2}{5} – (\frac{25}{y})^\frac{8}{5} = 0……. (2)
y^\frac{2}{5} – (\frac{5^2}{y})^\frac{8}{5} = 0
y = ± 5
Now, compare the roots -x1, + x2, ± y
It means, we cannot find any relation between x and y
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