Page 21 - mathsvol1ch1to3ans

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e
2
Solution: The equation x − ex + f = 0 has equal roots. The roots are .
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e
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The Product of the roots = = f. Since this root satisfy the first equation,
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e e e
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we have − a + b = 0. That is, f − a + b = 0. Hence ae = 2(b + f).
2 2 2
2
2
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8. Discuss the nature of roots of (i) −x + 3x + 1 = 0, (ii) 4x − x − 2 = 0, (iii) 9x + 5x = 0.
Solution:
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Equation b − 4ac Nature of the roots
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(i) −x + 3x + 1 = 0 9 + 4 = 13 Real and distinct
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(ii) 4x − x − 2 = 0 1 + 32 = 33 Real and distinct
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(iii) 9x + 5x = 0 25 + 0 = 25 Real and distinct
9. Without sketching the graphs, find whether the graphs of the following functions will intersect the
x-axis and if so in how many points.
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2
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(i) y = x + x + 2, (ii) y = x − 3x − 7, (iii) y = x + 6x + 9.
Solution:
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(i) x + x + 2 = 0 gives no real solution. Hence the function will not intersect the x axis.
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(ii)x −3x−7 = 0 gives 2 distinct values. Hence the function will intersect the x axis in 2 points.
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(iii)x + 6x + 9 = 0 gives 2 equal values. Hence the function will touch the x axis at 1 point.
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10. Write f(x) = x + 5x + 4 in completed square form.
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Solution:f(x) = x + 5x + 4
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f(x) = (x + 5x) + 4

25 25
2
= x + 5x + − + 4
4 4

25 9
2
= x + 5x + −
4 4
2 2
5 3
= x + −
2 2
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11. Find all values of m so that the quadratic function g(x) = (m − 2)x + 8x + m + 4 is negative
for all real values of x.
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Solution:g(x) = (m − 2)x + 8x + m + 4 < 0
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The quadratic function is negative for all real values of x if coefficient of x is negative. Hence
the values of m should be less than 2.

Exercise - 2.10

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1. Solve for x (i) 2x + x − 15 ≤ 0. (ii) −x + 3x − 2 ≥ 0.
Solution:

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(i) 2x + x − 15 ≤ 0. On factorizing the polynomial we get 2 (x + 3) x − ≤ 0
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