SATHEE: Theory of Equations 4 Question 8

Theory of Equations 4 Question 8

9. If $x^{2} + (a - b) x + (1 - a - b) = 0$ where $a, b \in R$ , then find the values of $a$ for which equation has unequal real roots for all values of $b$ .

$(2003, 4 M)$

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Answer:

Correct Answer: 9. A > 1

Solution:

Given,

$x^{2} + (a - b) x + (1 - a - b) = 0$ has real and unequal roots.

$\Rightarrow D > 0$

$\Rightarrow (a - b)^{2} - 4 (1) (1 - a - b) > 0$

$\Rightarrow a^{2} + b^{2} - 2 a b - 4 + 4 a + 4 b > 0$

Now, to find the values of ’ $a$ ’ for which equation has unequal real roots for all values of $b$ .

i.e. Above equation is true for all $b$ .

or $b^{2} + b (4 - 2 a) + (a^{2} + 4 a - 4) > 0$ , is true for all $b$ .

$∴$ Discriminant, $D < 0$

$\begin{array}{lrl} \Rightarrow & (4 - 2 a)^{2} - 4 (a^{2} + 4 a - 4) & < 0 \\ \Rightarrow & 16 - 16 a + 4 a^{2} - 4 a^{2} - 16 a + 16 & < 0 \\ \Rightarrow & - 32 a + 32 < 0 \Rightarrow a > 1 \end{array}$

Theory of Equations 4 Question 8

9. If $x^{2} + (a - b) x + (1 - a - b) = 0$ where $a, b \in R$ , then find the values of $a$ for which equation has unequal real roots for all values of $b$ .

Answer:

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Theory of Equations 4 Question 8

9. If x2+(a−b)x+(1−a−b)=0 where a,b∈R, then find the values of a for which equation has unequal real roots for all values of b.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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9. If $x^{2} + (a - b) x + (1 - a - b) = 0$ where $a, b \in R$ , then find the values of $a$ for which equation has unequal real roots for all values of $b$ .