Quadratic Expressions (VSAQs)

Maths-2A | 3. Quadratic Expressions – VSAQs:
Welcome to VSAQs in Chapter 3: Quadratic Expressions. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Find the nature of the roots of 3x²+7x+2=0

Here $$a = 3, b = 7, c = 2$$

$$\Delta = b^2 – 4ac = (7)^2 – 4(3)(2)$$

$$= 49 – 4(6) = 49 – 24 = 25 = 5^2 > 0$$

Here Δ is positive and a perfect square.

Hence, the roots are rational and unequal.

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VSAQ-2 : Prove that the roots of (x-a)(x-b) = h²

The equation is $$(x-a)(x-b) = h^2$$

$$\Rightarrow x^2 – (a + b)x + ab – h^2 = 0$$

Now, $$\Delta = (a + b)^2 – 4(ab – h^2)$$

$$= (a + b)^2 – 4ab + 4h^2 = (a – b)^2 + (2h)^2 > 0$$

Here Δ is positive.

Hence, the roots are real.

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VSAQ-3 : Form a quadratic equation, whose roots are 7 ± 2√5

We take $$\alpha = 7 + 2\sqrt{5}$$ and $$\beta = 7 – 2\sqrt{5}$$

$$\Rightarrow \alpha + \beta = (7 + 2\sqrt{5}) + (7 – 2\sqrt{5}) = 14$$

$$\alpha\beta = (7 + 2\sqrt{5})(7 – 2\sqrt{5}) = 49 – 20 = 29$$

The quadratic equation with roots α,β is $$x^2 – (\alpha + \beta)x + \alpha\beta = 0$$

$$\Rightarrow x^2 – 14x + 29 = 0$$

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VSAQ-4 : Form a quadratic equation, whose roots are (-3±5i)

Let $$\alpha = -3 + 5i$$ and $$\beta = -3 – 5i$$

$$\Rightarrow \alpha + \beta = (-3 + 5i) + (-3 – 5i) = -6$$

$$\alpha\beta = (-3 + 5i)(-3 – 5i) = 9 + 25 = 34$$

The quadratic equation with roots α,β is $$x^2 – (\alpha + \beta)x + \alpha\beta = 0$$

$$\Rightarrow x^2 + 6x + 34 = 0$$

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VSAQ-5 : If α,β are the roots of the equation ax²+bx+c=0, find the value of 1/α+1/β

Given α,β are the roots of $$ax^2 + bx + c = 0$$

$$\Rightarrow \text{Sum of the roots } \alpha + \beta = -\frac{b}{a}$$

$$\text{Product of the roots } \alpha\beta = \frac{c}{a}$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b}{a} \cdot \frac{a}{c} = -\frac{b}{c}$$

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VSAQ-6 : If α,β are the roots of the equation ax²+bx+c= 0 then find 1/α² +1/β²

Given α,β are roots of $$ax^2 + bx + c = 0$$

$$\Rightarrow \alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{(\alpha + \beta)^2 – 2\alpha\beta}{(\alpha\beta)^2}$$

$$= \left(\frac{-b}{a}\right)^2 – \frac{2\left(\frac{c}{a}\right)}{\left(\frac{c}{a}\right)^2}$$

$$= \frac{b^2}{a^2} – \frac{2c}{a} \cdot \frac{a^2}{c^2}$$

$$= \frac{b^2}{a^2} – \frac{2ac}{c^2}$$

$$= \frac{b^2 – 2ac}{c^2}$$

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VSAQ-7 : Find the maximum or minimum value of the expression x²-x+7

Comparing $$x^2 – x + 7$$ with $$ax^2 + bx + c$$ we get $$a = 1$$ $$b = -1$$ $$c = 7$$

Here, $$a = 1 > 0$$

So, we get the minimum value.

The minimum value is

$$\frac{4ac – b^2}{4a} = \frac{4(1)(7) – (-1)^2}{4(1)} = \frac{28 – 1}{4} = \frac{27}{4}$$

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VSAQ-8 : Find the maximum or minimum value of 2x-7-5x² where x∈R

Comparing $$2x – 7 – 5x^2$$ with $$ax^2 + bx + c$$ we get $$a = -5$$ $$b = 2$$ $$c = -7$$

Here, $$a = -5 < 0$$

So, we get the maximum value.

The maximum value is

$$\frac{4ac – b^2}{4a} = \frac{4(-5)(-7) – 2^2}{4(-5)} = \frac{140 – 4}{-20} = \frac{136}{-20} = -\frac{34}{5}$$

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VSAQ-9 : Find the values of m, if the equation x²-15-m(2x-8)=0 have equal roots

Given equation is $$x^2 – 15 – m(2x – 8) = 0$$

$$\Rightarrow x^2 – 15 – 2mx + 8m = 0$$

$$\Rightarrow x^2 – 2mx + (8m – 15) = 0$$

Comparing with $$ax^2 + bx + c = 0$$ we get

$$a = 1$$

$$b = -2m$$

$$c = 8m – 15$$

But roots are equal,

$$\Delta = b^2 – 4ac = 0$$

$$\Rightarrow (-2m)^2 – 4(1)(8m – 15) = 0$$

$$\Rightarrow 4m^2 – 32m + 60 = 0$$

$$\Rightarrow 4(m^2 – 8m + 15) = 0$$

$$\Rightarrow m^2 – 8m + 15 = 0$$

$$\Rightarrow m^2 – 5m – 3m + 15 = 0$$

$$\Rightarrow m(m-5) – 3(m-5) = 0$$

$$\Rightarrow (m-3)(m-5) = 0$$

$$\Rightarrow m = 3, 5$$

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VSAQ-10 : For what values of m. x²+(m+3)x+(m+6)=0 will have equal roots?

Given equation is $$x^2 + (m+3)x + (m+6) = 0$$

Comparing with $$ax^2 + bx + c = 0$$ we get

$$a = 1$$

$$b = m + 3$$

$$c = m + 6$$

But roots are equal,

$$\Delta = b^2 – 4ac = 0$$

$$\Rightarrow (m+3)^2 – 4(1)(m+6) = 0$$

$$\Rightarrow m^2 + 6m + 9 – 4m – 24 = 0$$

$$\Rightarrow m^2 + 2m – 15 = 0$$

$$\Rightarrow m^2 + 5m – 3m – 15 = 0$$

$$\Rightarrow m(m+5) – 3(m+5) = 0$$

$$\Rightarrow (m-3)(m+5) = 0$$

$$\Rightarrow m = 3, -5$$

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VSAQ-11 : If x²-6x+5=0 and x²-12x+p=0 have a common root then find p

Let α be the common root.

Then, $$\alpha^2 – 6\alpha + 5 = 0$$ and $$\alpha^2 – 12\alpha + p = 0$$

$$\Rightarrow (\alpha-1)(\alpha-5) = 0$$

$$\Rightarrow \alpha = 1 \text{ or } \alpha = 5$$

Put $$\alpha = 1$$ in $$\alpha^2 – 12\alpha + p = 0$$

$$\Rightarrow 1^2 – 12(1) + p = 0$$

$$\Rightarrow p = 12 – 1 = 11$$

Put $$\alpha = 5$$ in $$\alpha^2 – 12\alpha + p = 0$$

$$\Rightarrow 5^2 – 12(5) + p = 0$$

$$\Rightarrow 25 – 60 + p = 0$$

$$\Rightarrow p = 60 – 25 = 35$$

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VSAQ-12 : If the equations x²-6x+5=0, x²-3ax+35=0 have a common root then find a

Let α be the common root.

Then, $$\alpha^2 – 6\alpha + 5 = 0$$ and $$\alpha^2 – 3a\alpha + 35 = 0$$

$$\Rightarrow (\alpha-1)(\alpha-5) = 0$$

$$\Rightarrow \alpha = 1 \text{ or } \alpha = 5$$

Put $$\alpha = 1$$ in $$\alpha^2 – 3a\alpha + 35 = 0$$

$$\Rightarrow 1^2 – 3a(1) + 35 = 0$$

$$\Rightarrow 3a = 36$$

$$\Rightarrow a = 12$$

Put $$\alpha = 5$$ in $$\alpha^2 – 3a\alpha + 35 = 0$$

$$\Rightarrow 5^2 – 3a(5) + 35 = 0$$

$$\Rightarrow 25 – 15a + 35 = 0$$

$$\Rightarrow 15a = 60$$

$$\Rightarrow a = 4$$

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VSAQ-13 : If the equations x²+bx+c=0, x²+cx+b=0 have a common root then S.T b+c+1=0

Let α be the common root. Then

$$\alpha^2 + b\alpha + c = 0$$

$$\alpha^2 + c\alpha + b = 0$$

$$\alpha^2 + b\alpha + c – (\alpha^2 + c\alpha + b) = 0$$

$$\Rightarrow b\alpha – c\alpha + c – b = 0$$

$$\Rightarrow \alpha(b – c) = b – c$$

$$1^2 + b(1) + c = 0$$

$$\Rightarrow 1 + b + c = 0$$

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VSAQ-14 : If the equations x²+ax+b=0 and x²+cx+d=0 have a common root and the first equation has equal roots then prove that 2(b+d)=ac

Let α be the common root then

$$\alpha^2 + a\alpha + b = 0 \quad \text{(1)}$$

$$\alpha^2 + c\alpha + d = 0 \quad \text{(2)}$$

Also, since $$\alpha^2 + a\alpha + b = 0$$ has equal roots:

$$\Delta = 0 \Rightarrow a^2 – 4b = 0$$

$$\alpha + \alpha = -a \Rightarrow 2\alpha = -a \Rightarrow \alpha = -\frac{a}{2}$$

$$\alpha \cdot \alpha = b \Rightarrow \alpha^2 = b$$

Put these values in equation (2):

$$b + c\left(-\frac{a}{2}\right) + d = 0$$

$$\Rightarrow b + d = \frac{ac}{2}$$

$$\Rightarrow 2(b + d) = ac$$