Problem 21 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a<0\) the positive root of the equation \(x^{2}-2 a|x-a|\) \(-3 a^{2}=0\) is (A) \(a(-1-\sqrt{6})\) (B) \(a(-1+\sqrt{6})\) (C) \(a(1-\sqrt{2})\) (D) None of these

Step-by-Step Solution

Answer

Option (B): \(a(-1+\sqrt{6})\).

Step 1: Analyze the Absolute Value

Given the equation \(x^{2}-2a|x-a|-3a^{2}=0\), we know \(a<0\), so \(-a>0\). Since we need the positive root, consider two cases for the absolute value expression \(|x-a|\): \(x-a\) if \(x\geq a\) and \(-(x-a)\) if \(x

Step 2: Consider Case 1: \(x \geq a\)

If \(x \geq a\), then \(|x-a| = x-a\). Substitute into the equation: \(x^{2} - 2a(x-a) - 3a^{2} = 0\). Simplify to get \(x^{2} - 2ax + 2a^{2} - 3a^{2} = 0\), which simplifies further to: \(x^{2} - 2ax - a^{2} = 0\).

Step 3: Solve the Quadratic Equation (Case 1)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(x^{2} - 2ax - a^{2} = 0\). Here, \(a=1\), \(b=-2a\), \(c=-a^{2}\). So, \(x = \frac{-(-2a) \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot (-a^{2})}}{2 imes 1}\).

Step 4: Simplify the Quadratic Formula (Case 1)

Calculate further for \(x = \frac{2a \pm \sqrt{4a^{2} + 4a^{2}}}{2}\). This simplifies to \(x = \frac{2a \pm \sqrt{8a^{2}}}{2}\). So, \(x = \frac{2a \pm 2a\sqrt{2}}{2}\), simplifying to \(x = a(1 \pm \sqrt{2})\). Since we want the positive root, choose \(x = a(1 + \sqrt{2})\).

Step 5: Consider Case 2: \(x < a\)

If \(x < a\), then \(|x-a| = -(x-a) = -x + a\). Substitute into the original equation: \(x^{2} - 2a(-x+a) - 3a^{2} = 0\). This simplifies to \(x^{2} + 2ax - 2a^{2} - 3a^{2} = 0\), which becomes \(x^{2} + 2ax - 5a^{2} = 0\).

Step 6: Solve the Quadratic Equation (Case 2)

Using the quadratic formula for \(x^{2} + 2ax - 5a^{2} = 0\), we have \(a = 1\), \(b = 2a\), \(c = -5a^{2}\). Compute \(x = \frac{-2a \pm \sqrt{(2a)^2 - 4\cdot 1 \cdot (-5a^{2})}}{2}\).

Step 7: Simplify the Quadratic Formula (Case 2)

Calculate \(x = \frac{-2a \pm \sqrt{4a^{2} + 20a^{2}}}{2}\), which simplifies to \(x = \frac{-2a \pm \sqrt{24a^{2}}}{2}\). Simplify further to \(x = \frac{-2a \pm 2a\sqrt{6}}{2}\), or \(x = a(-1 \pm \sqrt{6})\). Since we need the positive root, choose \(x = a(-1 + \sqrt{6})\).

Step 8: Compare Solutions to Choices

The positive root found in Case 1 does not match any given options, while the positive root from Case 2 matches option (B): \(a(-1+\sqrt{6})\).

Key Concepts

Absolute Value

When solving quadratic equations, you might come across expressions with absolute values. An absolute value of a number refers to its distance from zero on the number line, regardless of direction.
This means it's always a non-negative number. For example,

• The absolute value of 3 is 3.
• The absolute value of -3 is also 3.

The function \( |x| \) can be expanded based on the sign of \(x\):

• If \(x \geq 0\), then \( |x| = x \).
• If \(x < 0\), then \( |x| = -x \).

In the presented quadratic equation, we handle the absolute value by considering these two scenarios and solving separately for each case.

Quadratic Formula

The quadratic formula is an essential tool for solving any quadratic equation. A quadratic equation generally looks like \( ax^2 + bx + c = 0 \). To solve it, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides both the possible roots, separated by the \"+\" and \"-\". Here, \( a \), \( b \), and \( c \) are coefficients from the equation.
Using the quadratic formula requires calculating the discriminant, \( b^2 - 4ac \). This value helps determine the nature of the roots:

• When the discriminant is positive, there are two real roots.
• When it is zero, there is one real root (a repeated root).
• If it's negative, the roots are complex.

In our given problem, applying this formula allows simplifying our equation to find possible values of \( x \).

Roots of Equations

Roots of a quadratic equation are the values of \( x \) where the equation equals zero. Simply put, they are solutions to the equation.
Every quadratic equation can have:

• Two distinct real roots.
• One repeated real root, when the roots are the same.
• Or two complex roots.

Finding these roots using conditions of the exercise, we first consider different scenarios defined by the absolute value, \( |x - a| \). Solving these separate cases and using the quadratic formula, we can derive at potential roots. Each approach may yield different sets or duplications of roots, but only one will ultimately satisfy all conditions from the exercise to be the correct solution.

Previous

Problem 20

Next

Problem 22

Other exercises in this chapter

Problem 19

If \(a, b, c \in R\) and quadratic equation \(x^{2}+(a+b) x+\) \(c=0\) has no real roots then (A) \(c(a+b+c)>0\) (B) \(c+c(a+b+c)>0\) (C) \(c+c(a+b-c)>0\) (D) \

View solution

Problem 20

If \(a x^{2}+b x+c=0, a \neq 0, a, b, c \in R\) has distinct real roots in \((1,2)\) then \(a\) and \(5 a+2 b+c\) have (A) same sign (B) opposite sign (C) not d

View solution

Problem 22

If \(p x^{2}+q x+r=0\) has no real roots and \(p, q, r\) are real such that \(p+r>0\), then (A) \(p-q+r \leq 0\) (B) \(p+r \geq q\) (C) \(p+r=q\) (D) None of th

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Problem 23

Given \(L x^{2}-m x+5=0\) does not have two distinct real roots, the minimum value of \(5 l+m\) is (A) 5 (B) \(-5\) (C) 1 (D) \(-1\)

View solution

Key Concepts

Absolute Value

When solving quadratic equations, you might come across expressions with absolute values. An absolute value of a number refers to its distance from zero on the number line, regardless of direction.
This means it's always a non-negative number. For example,

• The absolute value of 3 is 3.
• The absolute value of -3 is also 3.

The function \( |x| \) can be expanded based on the sign of \(x\):

• If \(x \geq 0\), then \( |x| = x \).
• If \(x < 0\), then \( |x| = -x \).

In the presented quadratic equation, we handle the absolute value by considering these two scenarios and solving separately for each case.

Quadratic Formula

The quadratic formula is an essential tool for solving any quadratic equation. A quadratic equation generally looks like \( ax^2 + bx + c = 0 \). To solve it, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides both the possible roots, separated by the \"+\" and \"-\". Here, \( a \), \( b \), and \( c \) are coefficients from the equation.
Using the quadratic formula requires calculating the discriminant, \( b^2 - 4ac \). This value helps determine the nature of the roots:

• When the discriminant is positive, there are two real roots.
• When it is zero, there is one real root (a repeated root).
• If it's negative, the roots are complex.

In our given problem, applying this formula allows simplifying our equation to find possible values of \( x \).

Roots of Equations

Roots of a quadratic equation are the values of \( x \) where the equation equals zero. Simply put, they are solutions to the equation.
Every quadratic equation can have:

• Two distinct real roots.
• One repeated real root, when the roots are the same.
• Or two complex roots.

Finding these roots using conditions of the exercise, we first consider different scenarios defined by the absolute value, \( |x - a| \). Solving these separate cases and using the quadratic formula, we can derive at potential roots. Each approach may yield different sets or duplications of roots, but only one will ultimately satisfy all conditions from the exercise to be the correct solution.