Algebra Simple Test paper. A set of problems to test your understanding of key algebra concepts. [New]

🧮 Algebra Mastery Paper (Grade X) ✒️

A set of problems to test your understanding of key algebra concepts. Analyze each question, solve it step-by-step, and then verify your answer.

Question 1: Quadratic Equations ➕➖

Solve the quadratic equation \( 2x^2 + 5x - 12 = 0 \) using the quadratic formula.

✅ Solution

The given equation is \( 2x^2 + 5x - 12 = 0 \).

We compare this to the standard form \(ax^2 + bx + c = 0\) to get the coefficients:

• \(a = 2\)
• \(b = 5\)
• \(c = -12\)

The quadratic formula is:

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

Substitute the values of \(a, b,\) and \(c\):

$$ x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)} $$ $$ x = \frac{-5 \pm \sqrt{25 + 96}}{4} $$ $$ x = \frac{-5 \pm \sqrt{121}}{4} $$ $$ x = \frac{-5 \pm 11}{4} $$

Now, we find the two possible values for \(x\):

$$ x_1 = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2} $$ $$ x_2 = \frac{-5 - 11}{4} = \frac{-16}{4} = -4 $$

The solutions (roots) of the equation are 3/2 and -4. 💯

Question 2: Arithmetic Progressions 📈

The 4th term of an Arithmetic Progression (AP) is 11 and the 9th term is 21. Find the first term (\(a\)), the common difference (\(d\)), and the 15th term (\(a_{15}\)).

✅ Solution

The formula for the \(n\)-th term of an AP is \( a_n = a + (n-1)d \).

We are given:

• The 4th term is 11: \( a_4 = a + (4-1)d = a + 3d = 11 \) --- (Equation 1)
• The 9th term is 21: \( a_9 = a + (9-1)d = a + 8d = 21 \) --- (Equation 2)

To find \(d\), subtract Equation 1 from Equation 2:

$$ (a + 8d) - (a + 3d) = 21 - 11 $$ $$ 5d = 10 $$ $$ d = 2 $$

The common difference (\(d\)) is 2.

Now, substitute \(d=2\) into Equation 1 to find \(a\):

$$ a + 3(2) = 11 $$ $$ a + 6 = 11 $$ $$ a = 5 $$

The first term (\(a\)) is 5.

Finally, find the 15th term (\(a_{15}\)):

$$ a_{15} = a + (15-1)d = 5 + 14(2) $$ $$ a_{15} = 5 + 28 = 33 $$

The 15th term of the AP is 33. 🎯

Question 3: Polynomials 🔢

Find the zeroes of the quadratic polynomial \( p(x) = x^2 - 2x - 8 \) and verify the relationship between the zeroes and the coefficients.

✅ Solution

First, we find the zeroes by setting \(p(x) = 0\):

$$ x^2 - 2x - 8 = 0 $$

We can factor the quadratic by finding two numbers that multiply to -8 and add to -2. These numbers are -4 and 2.

$$ (x - 4)(x + 2) = 0 $$

This gives us the zeroes:

$$ x - 4 = 0 \implies x = 4 $$ $$ x + 2 = 0 \implies x = -2 $$

The zeroes are \(\alpha = 4\) and \(\beta = -2\).

---

Verification:

For the polynomial \(ax^2 + bx + c\), the coefficients are \(a=1, b=-2, c=-8\).

1. Sum of the zeroes (\(\alpha + \beta\)):

$$ \text{From zeroes:} \quad 4 + (-2) = 2 $$ $$ \text{From formula:} \quad -\frac{b}{a} = -\frac{-2}{1} = 2 $$

The sum is verified. ✅

2. Product of the zeroes (\(\alpha\beta\)):

$$ \text{From zeroes:} \quad 4 \times (-2) = -8 $$ $$ \text{From formula:} \quad \frac{c}{a} = \frac{-8}{1} = -8 $$

The product is verified. ✅

Question 4: Linear Equations Word Problem 💰

Five years ago, Neha was three times as old as Sonu. Ten years from now, Neha will be twice as old as Sonu. What are their present ages?

✅ Solution

Let Neha's present age be \(x\) years and Sonu's present age be \(y\) years.

Condition 1: Five years ago

• Neha's age: \(x - 5\)
• Sonu's age: \(y - 5\)

The equation is: \( x - 5 = 3(y - 5) \implies x - 5 = 3y - 15 \implies x - 3y = -10 \) --- (1)

Condition 2: Ten years from now

• Neha's age: \(x + 10\)
• Sonu's age: \(y + 10\)

The equation is: \( x + 10 = 2(y + 10) \implies x + 10 = 2y + 20 \implies x - 2y = 10 \) --- (2)

Now, we solve the system of linear equations. Subtract Equation (1) from Equation (2):

$$ (x - 2y) - (x - 3y) = 10 - (-10) $$ $$ y = 20 $$

Sonu's present age is 20 years.

Substitute \(y=20\) into Equation (2):

$$ x - 2(20) = 10 $$ $$ x - 40 = 10 $$ $$ x = 50 $$

Neha's present age is 50 years. 🧑‍🤝‍🧑

Well done! Algebra is a powerful tool, and practice is the key to unlocking its potential. 🚀