Class 10 Mathematics Chapter 4 Quadratic Equations

Key Concept: Coefficient Identification

c) a = 5, b = -8, c = 3
[Solution Description] Comparing the given equation with the standard form $ax^2 + bx + c = 0$, we find a = 5, b = -8, and c = 3.

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Key Concept: Simplify to Quadratic

a) Yes, it is a quadratic equation
[Solution Description] Start by expanding the right side: $2(1-x) = 2 – 2x$. Rewrite the equation as $x^2 – 3x = 2 – 2x$. Bringing all terms to one side gives $x^2 – 3x – 2 + 2x = 0$, which simplifies to $x^2 – x – 2 = 0$, a quadratic equation.

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Key Concept: Quadratic Formula Application

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] To determine the nature of the roots for the given quadratic equation $3x^2 – 6x + 3 = 0$, we calculate the discriminant $\Delta = b^2 – 4ac$. Here, $a = 3$, $b = -6$, and $c = 3$.

First, compute the discriminant: $\Delta = (-6)^2 – 4(3)(3)$

$\Delta = 36 – 36 = 0$

Since $\Delta = 0$, the quadratic equation has two equal roots.

The reason provided states that for any quadratic equation $ax^2 + bx + c = 0$, if $b^2 – 4ac = 0$, it indeed has two equal roots. This is true by definition of the discriminant in relation to the nature of the roots.

Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

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Key Concept: Real-life Application

b) Calculating the maximum height of a projectile
[Solution Description] Calculating the maximum height of a projectile involves quadratic equations since the motion of projectiles is governed by quadratic relationships due to gravity.

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Key Concept: Discriminant Identification

b) $b^2 – 4ac = 0$, one real root (repeated)
[Solution Description] The discriminant is given by $b^2 – 4ac$. For the equation $4x^2 + 12x + 9 = 0$, a = 4, b = 12, and c = 9. Calculate as follows:
$b^2 – 4ac = 12^2 – 4 \times 4 \times 9 = 144 – 144 = 0$
A discriminant of 0 indicates one real root (repeated).​​

Key Concept: Real Roots Existence

c) No, it does not have real roots
[Solution Description] The discriminant $\Delta$ is given by $= b^2 – 4ac =$. For the equation $= 3x^2 + x + 4 = 0 =$, we have a = 3, b = 1, and c = 4. Therefore, $\Delta = (1)^2 – 4 \cdot 3 \cdot 4 = 1 – 48 = -47.$
Since $\Delta < 0$, the equation does not have real roots. Click Here to Download Notes

Key Concept: Standard Form Recognition

d) $x^2 – 6x + 8 = 0$
[Solution Description] The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Option d) $x^2 – 6x + 8 = 0$ is in this form.

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Key Concept: Equal Roots Condition

a) $x^2 – 4x + 4 = 0$
[Solution Description] To have equal roots, the discriminant must be zero ($b^2 – 4ac = 0$).

Let’s check each option:

a) $x^2 – 4x + 4 = 0$: $D = (-4)^2 – 4 \cdot 1 \cdot 4 = 16 – 16 = 0$

b) $x^2 + 2x + 1 = 0$: $D = 2^2 – 4 \cdot 1 \cdot 1 = 4 – 4 = 0$

c) $x^2 – 2x + 3 = 0$: $D = (-2)^2 – 4 \cdot 1 \cdot 3 = 4 – 12 = -8$

d) $x^2 + 3x + 2 = 0$: $D = 3^2 – 4 \cdot 1 \cdot 2 = 9 – 8 = 1$

Both options a) and b) have equal roots as they have D = 0.

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Key Concept:Complex Factorization, Real-world Application

a) $w^2 + 5w – 84 = 0$

[Solution Description] Let w represent the width of the garden. Then the length of the garden is $w + 5$. The area of a rectangle is given by the product of its length and width. Thus, we have the equation: $w(w + 5) = 84$

Simplifying this gives: $w^2 + 5w – 84 = 0$

This is the quadratic equation that represents the problem.

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Key Concept: Equation Comparison

c) $2x^2 + 4x + 1 = 0$
[Solution Description] The characteristics of a quadratic equation include having terms up to $x^2$, with no higher degree terms. Option c) fits the description since it has an $x^2$ term and can be arranged to $ax^2 + bx + c = 0$ format.

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Key Concept: Real-life Scenario

a) $ w^2 + 3w – 40 = 0 $
[Solution Description] Let the width of the rectangle be w . Then, the length of the rectangle is w + 3 . The area of the rectangle is given by the formula Area = length $\times$ width. $w(w + 3) = 40$
Expanding and rearranging the equation gives: $w^2 + 3w – 40 = 0$
This is the quadratic equation representing the scenario.

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Key Concept: Factorization Method

b) x = 2, 3
[Solution Description] We need to factorize the equation $x^2 – 5x + 6 = 0$. Let’s find two numbers whose product is 6 (the constant term) and sum is -5 (the coefficient of x). The numbers are -2 and -3. Therefore,
\begin{equation*}
x^2 – 5x + 6 = (x – 2)(x – 3) = 0
\end{equation*}
Setting each factor to zero gives $x = 2$ and $x = 3$.

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Key Concept: Factorization Basics

a) (x + 2)(x + 3)
[Solution Description] The given equation is $x^2 + 5x + 6 = 0$. We need to split the middle term, which is 5x, into two terms whose product is equal to the product of the coefficient of $x^2$ (which is 1) and the constant term (6), that is $1 \times 6 = 6$.

We can express $5x$ as $2x + 3x$ because $2 \times 3 = 6$. So the equation becomes $x^2 + 2x + 3x + 6 = 0$.

Now, take common factors from the first two terms and last two terms:

– From $x^2 + 2x$, take $x$ common: $x(x + 2)$
– From $3x + 6$, take $3$ common: $3(x + 2)$

This gives us: $(x + 2)(x + 3) = 0$

Thus, the factorization is $(x + 2)(x + 3)$.

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Key Concept: Factorization Basics

b) (x + 2)(x + 3)
[Solution Description] To factorize the quadratic equation $x^2 + 5x + 6$, we look for two numbers that multiply to 6 (the constant term) and add up to 5 (the coefficient of the middle term). The numbers 2 and 3 satisfy these conditions: $2 \cdot 3 = 6$ and $2 + 3 = 5$. Thus, the factorization is $(x + 2)(x + 3)$.

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