Factorise:
Question: Factorise: $4 a^{2}-9 b^{2}-2 a-3 b$ Solution: $4 a^{2}-9 b^{2}-2 a-3 b$ $=(2 a)^{2}-(3 b)^{2}-1(2 a+3 b)$ $=(2 a+3 b)(2 a-3 b)-1(2 a+3 b) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(2 a+3 b)[(2 a-3 b)-1]$ $=(2 a+3 b)(2 a-3 b-1)$...
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Question: Factorise: $a^{2}-b^{2}-a-b$ Solution: $a^{2}-b^{2}-a-b$ $=(a+b)(a-b)-1(a+b) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$ $=(a+b)[(a-b)-1]$ $=(a+b)(a-b-1)$...
Read More →The values of k for which the quadratic equation
Question: The values of $k$ for which the quadratic equation $k x^{2}+1=k x+3 x-11 x^{2}$ has real and equal roots are (a) 11, 3 (b) 5, 7 (c) 5, 7 (d) none of these Solution: (c) 5, 7 The given equation is $k x^{2}+1=k x+3 x-11 x^{2}$ which can be written as. $k x^{2}+11 x^{2}-k x-3 x+1=$ $\Rightarrow(k+11) x^{2}-(k+3) x+1=0$ For equal and real roots, the discriminant of $(k+11) x^{2}-(k+3) x+1=0$. $\therefore(k+3)^{2}-4(k+11)=0$ $\Rightarrow k^{2}+2 k-35=0$ $\Rightarrow(k-5)(k+7)=0$ $\Rightarro...
Read More →The values of k for which the quadratic equation
Question: The values of $k$ for which the quadratic equation $k x^{2}+1=k x+3 x-11 x^{2}$ has real and equal roots are (a) 11, 3 (b) 5, 7 (c) 5, 7 (d) none of these Solution: (c) 5, 7 The given equation is $k x^{2}+1=k x+3 x-11 x^{2}$ which can be written as. $k x^{2}+11 x^{2}-k x-3 x+1=$ $\Rightarrow(k+11) x^{2}-(k+3) x+1=0$ For equal and real roots, the discriminant of $(k+11) x^{2}-(k+3) x+1=0$. $\therefore(k+3)^{2}-4(k+11)=0$ $\Rightarrow k^{2}+2 k-35=0$ $\Rightarrow(k-5)(k+7)=0$ $\Rightarro...
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Question: Factorise: $(3 a+5 b)^{2}-4 c^{2}$ Solution: $(3 a+5 b)^{2}-4 c^{2}$ $=(3 a+5 b)^{2}-(2 c)^{2}$ $=(3 a+5 b+2 c)(3 a+5 b-2 c) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →The value of a such that x
Question: The value of $a$ such that $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$ may have a common root is (a) 0 (b) 12 (c) 24 (d) 32 Solution: (a) and (c) Let $\alpha$ be the common roots of the equations $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$. Therefore, $\alpha^{2}-11 \alpha+a=0$ ...(1) $\alpha^{2}-14 \alpha+2 a=0$ ...(2) Solving (1) and (2) by cross multiplication, we get, $\frac{\alpha^{2}}{-22 a+14 a}=\frac{\alpha}{a-2 a}=\frac{1}{-14+11}$ $\Rightarrow \alpha^{2}=\frac{-22 a+14 a}{-14+11}, \a...
Read More →The value of a such that x
Question: The value of $a$ such that $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$ may have a common root is (a) 0 (b) 12 (c) 24 (d) 32 Solution: (a) and (c) Let $\alpha$ be the common roots of the equations $x^{2}-11 x+a=0$ and $x^{2}-14 x+2 a=0$. Therefore, $\alpha^{2}-11 \alpha+a=0$ ...(1) $\alpha^{2}-14 \alpha+2 a=0$ ...(2) Solving (1) and (2) by cross multiplication, we get, $\frac{\alpha^{2}}{-22 a+14 a}=\frac{\alpha}{a-2 a}=\frac{1}{-14+11}$ $\Rightarrow \alpha^{2}=\frac{-22 a+14 a}{-14+11}, \a...
Read More →Factorise:
Question: Factorise: $20 x^{2}-45$ Solution: $20 x^{2}-45$ $=5\left(4 x^{2}-9\right)$ $=5\left[(2 x)^{2}-3^{2}\right]$ $=5(2 x+3)(2 x-3) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
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Question: Factorise: $20 x^{2}-45$ Solution: $20 x^{2}-45$ $=5\left(4 x^{2}-9\right)$ $=5\left[(2 x)^{2}-3^{2}\right]$ $=5(2 x+3)(2 x-3) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →In the following, determine whether the given quadratic equations have real roots and if so, find the roots:
Question: In the following, determine whether the given quadratic equations have real roots and if so, find the roots: (i) $16 x^{2}=24 x+1$ (ii) $x^{2}+x+2=0$ (iii) $\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$ (iv) $3 x^{2}-2 x+2=0$ (v) $2 x^{2}-2 \sqrt{6} x+3=0$ (vi) $3 a^{2} x^{2}+8 a b x+4 b^{2}=0, a \neq 0$ (vii) $3 x^{2}+2 \sqrt{5} x-5=0$ (viii) $x^{2}-2 x+1=0$ (ix) $2 x^{2}+5 \sqrt{3} x+6=0$ (x) $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ (xi) $2 x^{2}-2 \sqrt{2} x+1=0$ (xii) $3 x^{2}-5 x+2=0$ Solution: In t...
Read More →If the roots of x
Question: If the roots of $x^{2}-b x+c=0$ are two consecutive integers, then $b^{2}-4 c$ is (a) 0 (b) 1 (c) 2 (d) none of these. Solution: (b) 1 Given equation: $x^{2}-b x+c=0$ Let $\alpha$ and $\alpha+1$ be the two consecutive roots of the equation. Sum of the roots $=\alpha+\alpha+1=2 \alpha+1$ Product of the roots $=\alpha(\alpha+1)=\alpha^{2}+\alpha$ So, sum of the roots $=2 \alpha+1=\frac{-\text { Coeffecient of } x}{\text { Coeffecient of } x^{2}}=\frac{b}{1}=b$ Product of the roots $=\alp...
Read More →If the roots of x
Question: If the roots of $x^{2}-b x+c=0$ are two consecutive integers, then $b^{2}-4 c$ is (a) 0 (b) 1 (c) 2 (d) none of these. Solution: (b) 1 Given equation: $x^{2}-b x+c=0$ Let $\alpha$ and $\alpha+1$ be the two consecutive roots of the equation. Sum of the roots $=\alpha+\alpha+1=2 \alpha+1$ Product of the roots $=\alpha(\alpha+1)=\alpha^{2}+\alpha$ So, sum of the roots $=2 \alpha+1=\frac{-\text { Coeffecient of } x}{\text { Coeffecient of } x^{2}}=\frac{b}{1}=b$ Product of the roots $=\alp...
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Question: Factorise: $2-50 x^{2}$ Solution: $2-50 x^{2}$ $=2\left(1-25 x^{2}\right)$ $=2\left[1^{2}-(5 x)^{2}\right]$ $=2(1+5 x)(1-5 x) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$...
Read More →If x is real and k
Question: If $x$ is real and $k=\frac{x^{2}-x+1}{x^{2}+x+1}$, then (a)k [1/3,3] (b)k 3 (c)k 1/3 (d) none of these Solution: (a) $k \in[1 / 3,3]$ $k=\frac{x^{2}-x+1}{x^{2}+x+1}$ $\Rightarrow k x^{2}+k x+k=x^{2}-x+1$ $\Rightarrow(k-1) x^{2}+(k+1) x+k-1=0$ For real values of $x$, the discriminant of $(k-1) x^{2}+(k+1) x+k-1=0$ should be greater than or equal to zero. $\therefore$ if $k \neq 1$ $(k+1)^{2}-4(k-1)(k-1) \geq 0$ $\Rightarrow(k+1)^{2}-\{2(k-1)\}^{2} \geq 0$ $\Rightarrow(k+1+2 k-2)(k+1-2 ...
Read More →Factorize:
Question: Factorize: $150-6 x^{2}$ Solution: $150-6 x^{2}=6\left(25-x^{2}\right)$ $=6\left(5^{2}-x^{2}\right)$ $=6(5-x)(5+x)$...
Read More →The number of solutions of x
Question: The number of solutions of $x^{2}+|x-1|=1$ is (a) 0 (b) 1 (c) 2 (d) 3 Solution: (c) 2 $x^{2}+|x-1|=x^{2}+x-1, x \geq 1$ $=x^{2}-x+1 \quad, x1$ (i)$x^{2}+x-1=1$ $\Rightarrow x^{2}+x-2=0$ $\Rightarrow x^{2}+2 x-x-2=0$ $\Rightarrow x(x+2)-1(x+2)=0$ $\Rightarrow(x+2)(x-1)=0$ $\Rightarrow x+2=0$ or, $x-1=0$ $\Rightarrow x=-2$ or $x=1$ Since $-2$ does not satisfy the condition $x \geq 1$ (ii)$x^{2}-x+1=1$ $\Rightarrow x^{2}-x=0$ $\Rightarrow x^{2}-x=0$ $\Rightarrow x(x-1)=0$ $\Rightarrow x=0...
Read More →The number of solutions of x
Question: The number of solutions of $x^{2}+|x-1|=1$ is (a) 0 (b) 1 (c) 2 (d) 3 Solution: (c) 2 $x^{2}+|x-1|=x^{2}+x-1, x \geq 1$ $=x^{2}-x+1 \quad, x1$ (i)$x^{2}+x-1=1$ $\Rightarrow x^{2}+x-2=0$ $\Rightarrow x^{2}+2 x-x-2=0$ $\Rightarrow x(x+2)-1(x+2)=0$ $\Rightarrow(x+2)(x-1)=0$ $\Rightarrow x+2=0$ or, $x-1=0$ $\Rightarrow x=-2$ or $x=1$ Since $-2$ does not satisfy the condition $x \geq 1$ (ii)$x^{2}-x+1=1$ $\Rightarrow x^{2}-x=0$ $\Rightarrow x^{2}-x=0$ $\Rightarrow x(x-1)=0$ $\Rightarrow x=0...
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Question: Factorize: $8 a b^{2}-18 a^{3}$ Solution: $8 a b^{2}-18 a^{3}=2 a\left(4 b^{2}-9 a^{2}\right)$ $=2 a\left[(2 b)^{2}-(3 a)^{2}\right]$ $=2 a(2 b-3 a)(2 b+3 a)$...
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Question: Factorize: $x-64 x^{3}$ Solution: $x-64 x^{3}=x\left(1-64 x^{2}\right)$ $=x\left[1-(8 x)^{2}\right]$ $=x(1-8 x)(1+8 x)$...
Read More →The number of real solutions of
Question: The number of real solutions of $\left|2 x-x^{2}-3\right|=1$ is (a) 0 (b) 2 (c) 3 (d) 4 Solution: (b) 2 Given equation: $\left|2 x-x^{2}-3\right|=1$ (i)$2 x-x^{2}-3=1$ $\Rightarrow 2 x-x^{2}-4=0$ $\Rightarrow x^{2}-2 x+4=0$ $\Rightarrow(x-2)^{2}=0$ $\Rightarrow x=2,2$ (ii)$-2 x+x^{2}+3=1$ $\Rightarrow x^{2}-2 x+2=0$ $\Rightarrow x^{2}-2 x+1+1=0$ $\Rightarrow(x-1)^{2}-i^{2}=0$ $\Rightarrow(x-1+i)(x-1-i)=0$ $\Rightarrow x=1-i, 1+i$ Hence, the real solutions are 2,2 ....
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Question: Factorize: $27 a^{2}-48 b^{2}$ Solution: $27 a^{2}-48 b^{2}=3\left(9 a^{2}-16 b^{2}\right)$ $=3\left[(3 a)^{2}-(4 b)^{2}\right]$ $=3(3 a-4 b)(3 a+4 b)$...
Read More →The number of real solutions of
Question: The number of real solutions of $\left|2 x-x^{2}-3\right|=1$ is (a) 0 (b) 2 (c) 3 (d) 4 Solution: (b) 2 Given equation: $\left|2 x-x^{2}-3\right|=1$ (i)$2 x-x^{2}-3=1$ $\Rightarrow 2 x-x^{2}-4=0$ $\Rightarrow x^{2}-2 x+4=0$ $\Rightarrow(x-2)^{2}=0$ $\Rightarrow x=2,2$ (ii)$-2 x+x^{2}+3=1$ $\Rightarrow x^{2}-2 x+2=0$ $\Rightarrow x^{2}-2 x+1+1=0$ $\Rightarrow(x-1)^{2}-i^{2}=0$ $\Rightarrow(x-1+i)(x-1-i)=0$ $\Rightarrow x=1-i, 1+i$ Hence, the real solutions are 2,2 ....
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Question: Factorize: $3 x^{3}-48 x$ Solution: $3 x^{3}-48 x=3 x\left(x^{2}-16\right)$ $=3 x\left(x^{2}-4^{2}\right)$ $=3 x(x-4)(x+4)$...
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}+p x+1=0 ; \gamma, \delta$ the roots of the equation $x^{2}+q x+1=0$, then $(\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta)=$ (a) $q^{2}-p^{2}$ (b) $p^{2}-q^{2}$ (c) $p^{2}+q^{2}$ (d) none of these. Solution: (a) $q^{2}-p^{2}$ Given: $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+1=0$. Also, $\gamma$ and $\delta$ are the roots of the equation $x^{2}+q x+1=0$. Then, the sum and the product of the roots of t...
Read More →If α, β are the roots of the equation x
Question: If $\alpha, \beta$ are the roots of the equation $x^{2}+p x+1=0 ; \gamma, \delta$ the roots of the equation $x^{2}+q x+1=0$, then $(\alpha-\gamma)(\alpha+\delta)(\beta-\gamma)(\beta+\delta)=$ (a) $q^{2}-p^{2}$ (b) $p^{2}-q^{2}$ (c) $p^{2}+q^{2}$ (d) none of these. Solution: (a) $q^{2}-p^{2}$ Given: $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+1=0$. Also, $\gamma$ and $\delta$ are the roots of the equation $x^{2}+q x+1=0$. Then, the sum and the product of the roots of t...
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