>> Elementary Mathematics >> Quadratic Equations

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41). If one root of the equation \( \Large x^{2}+px+12=0 \) is 4, while the equation \( \Large x^{2}-7x+q=0 \) has equal roots, then the value of 'g' is:
Since, 4 is one the roots of equation \( \Large x^{2}+px+12=0 \) so it must satisfies the equation | ||||

42). Let \( \Large 2\sin^{2}+3\sin x-2>0 \) and \( \Large x^{2}-x-2<0 \) (x is measured in radians). Then x lies in the interval:
Given equation is | ||||

43). If at least one root of \( \Large 2x^{2}+3x+5=0 \) and \( \Large ax^{2}+bx+c=0 \), a, b, c, belongs to N is common, then the maximum value of a + b + c is:
Roots of the equation \( \Large 2x^{2}+3x+5=0 \) \( \Large x = \frac{-3\pm \sqrt{9-40}}{6} \)( imaginary roots). Hence, both roots coincide, so on comparing \( \Large \frac{a}{2}=\frac{b}{3}=\frac{c}{5}=K \) => \( \Large a=2K,\ b=3K,\ c=5K \) => \( \Large a+b+c = 10K \) So, maximum value does not exist. | ||||

44). If the roots of the quadratic equation \( \Large x^{2}+px+q=0 \) are \( \Large \tan 30 ^{\circ} and\ \tan 15 ^{\circ} \) respectively, then the value of\( \Large 2+q-p \) is
Since, \( \Large \tan 30 ^{\circ} and\ \tan 15 ^{\circ} \) are roots of equation \( \Large x^{2}+px+q=0 \) \( \Large \tan 30 ^{\circ} + \tan 15 ^{\circ} = -P \) and \( \Large \tan 30 ^{\circ} \tan 15 ^{\circ} = -q \) \( \Large \therefore 2+q-P=2+\tan 30 ^{\circ} + \tan 15 ^{\circ} + \left(\tan 30 ^{\circ} + \tan 15 ^{\circ} \right) \) =\( \Large 2+\tan 30 ^{\circ} \tan 15 ^{\circ} +1- \tan 30 ^{\circ} \tan 15 ^{\circ} \) \( \Large \left(\because \tan 45 ^{\circ} =\frac{\tan 30 ^{\circ} + \tan 15 ^{\circ} }{1- \tan 30 ^{\circ} \tan 15 ^{\circ} }\right)=2+q-P=3 \) | ||||

45). If \( \Large P \left(x\right)=ax^{2}+bx+c \) and \( \Large Q \left(x\right)=-ax^{2}+dx+c \) where \( \Large ac \ne 0 \) then \( \Large P \left(x\right) Q \left(x\right) = 0 \) has at least:
Let all four roots are imaginary. Then roots of both equation \( \Large P \left(x\right)=0 \) | ||||

46). The coefficient of x in the equation \( \Large x^{2}+px+q=0 \) was taken as 17 in place of 13 its roots Were found to be -2 and -15. The roots of the original equation are:
Let the equation (incorrectly written form) be \( \Large x^{2}+17x+q=0 \) since, roots are -2, -15 Therefore, q = 30 so, correct equation is \( \Large x^{2}+13x+30=0 \) => \( \Large x^{2}+10x+3x+30=0 \) => \( \Large \left(x+3\right) \left(x+10\right)=0 \) => x = -3, -10 | ||||

47). The number which exceeds its positive Square roots by 12 is:
Let the required number is x According to given condition \( \Large x=\sqrt{x}+12 => x-12=\sqrt{x} \) => \( \Large x^{2}-25x+144=0 \) => \( \Large x^{2}-16x-9x+144=0 \) Therefore, x = 16, 9 since x = 9 does not hold the condition Therefore, x = 16 | ||||

48). Let a, b, c be real numbers a \( \ne \) 0. If \( \Large \alpha \) is a root of \( \Large a^{2}x^{2}+bx+c=0, \),\( \Large \beta \) is a root of \( \Large a^{2}x^{2}-bx-c=0 \) and \( \Large 0< \alpha < \beta \) then the equation \( \Large a^{2}x^{2}+2bx+2c=0 \) has a root of \( \gamma \) that always satisfies:
Since, \( \Large \alpha \) and \( \Large \beta \) are the roots of given equation | ||||

49). The equation \( \Large x \left(\frac{3}{4}log_{2}x\right)^{2}+ \left(log_{2}x\right) -\frac{5}{4}=\sqrt{2} \) has
For given equation to be meaningful we must have x > 0 for x > 0, the given equation can be written as \( \Large \frac{3}{4} \left(log_{2}x\right)^{2}+log_{2}x-\frac{5}{4} \) | ||||

50). The solution of set of the equation \( \Large x log x \left(1-x\right)^{2}=9 \) is
We have, \( \Large x^{logx \left(1-x\right)^{2} }=9 \) Taking log on both sides, we get \( \Large x \left(9\right)=log x \left(1-x\right)^{2} \left(\because a^{x}=N => log a N = x \right) \) => \( \Large 9= \left(1-x\right)^{2} => 1+x^{2}-2x-9=0 \) => \( \Large x^{2}-2x-8=0 \) => \( \Large x = -2, 4 \) => \( \Large x = 4 \left(\because x = 2\right) \) |