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This is the second of our many quadratic formula quizzes. Click on this to access more such competitive quizzes on quadratic equations.

Quadratic Formula Quiz 2

mathematics

 A flower bouquet is made by arranging tulips and myrtle leaves. The number of tulips is 5 more than the number of myrtle leaves. The product of the number of tulips and myrtle leaves is 25. What would be the number of tulips that can be arranged in the bouquet?

 Form the quadratic equation to find the number of myrtle leaves. Then add 5 to the obtained result to get the answer.

 $\text{Take the number of myrtle leaves as “x”.}\\ \text{So, the number of tulips would be "5+x”.}\\ \text{The product of number of tulips and myrtle leaves = 150} \\ \Rightarrow x\left( 5+x \right)=150 \\ \Rightarrow {{x}^{2}}+5x-150=0 \\ \text{Using quadratic formula, } x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times (-150)}}{2} \\ \Rightarrow x=\dfrac{-5\pm \sqrt{25+600}}{2} \\ \Rightarrow x=\dfrac{-5\pm 25}{2} \\ \Rightarrow x=10,-15 \\ \therefore \text{Taking the positive value, the least number of tulips that can be arranged would be }\\ 10 + 5 = 15$.

 Consider the weighing scale as shown below. There are 3 yellow marbles weighing $2gm$ each. There are some pink marbles which need to be placed on the second plate. The number of pink marbles is equal to the weight of each pink marble. If the total weight of both yellow marbles and pink marbles is $31gm$, then choose the correct relation from the options given below. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fc171c17-4947-4239-9d25-69c1490ed1444773330670182542116.png"/>

 Assume the number of pink marble as $x$ and its weight is $x gm$. Then by adding the weight of both marbles, equate them. Lastly deduce the relation between the weight of both marbles.

 $\text{Weight of 3 yellow marbles = }3\times 2gm=6gm.\\ \text{Let the number of pink marbles be “x”.}\\ \text{So, weight of “x” pink marbles = }x\times x gm={{x}^{2}}gm.\\ \text{Total weight is 31gm, so we have } 6+{{x}^{2}}=31 \\ \therefore x=5 \\ \text{So, there are 5 pink marbles weighing 5gm each.} \\ \text{Now we have Total weight of yellow marbles = 6gm.} \\ \text{Total weight of pink marbles = 25gm}.\\ \text{Hence, Weight of yellow marbles } < \text{ Weight of pink marbles}$

 Find the value of h if ${{h}^{3}}-{{\left( h-1 \right)}^{3}}=2$.

 Use the expansion for ${{\left( a-b \right)}^{3}}$ and convert the given equation into a quadratic equation. Then apply the quadratic formula and obtain the value of $h$.

 $\text{Convert the given equation in the form }a{{x}^{2}}+bx+c=0 \\ \text{Apply the quadratic formula }{{h}^{3}}-{{\left( h-1 \right)}^{3}}=2 \\ \Rightarrow 3{{h}^{2}}-3h-1=0 \\ \Rightarrow h=\dfrac{-(-3)\pm \sqrt{{{(-3)}^{2}}-4\times 3\times -1}}{2\times 3} \\ \Rightarrow h=\dfrac{3\pm \sqrt{21}}{6} \\ \therefore h=\dfrac{1}{2}\pm \dfrac{\sqrt{21}}{6} $ 

 There are four children in a family. The eldest child is twice as old as the pair of twins. The youngest child is 6 years younger than the elder one. If the sum of ages of the eldest child, the youngest child and the product of ages of the pair of twins is 26, then find the age of the eldest one.

 Take the age of the twins as the unknown variable. Find the age of the eldest and youngest child in terms of $x$. Add the ages as per the condition and form a quadratic equation in $x$. Solve for $x$ and then find the eldest child’s age using the relation given in the question.

 $\text{Let the age of one twin be “x” years.} \\ \text{So, the age of the other twin will also be “x” years.} \\ \text{Now, the age of the eldest child }= 2x \\ \text{Age of the youngest child }= 2x – 6 \\ \text{Sum of ages}=2x+{x^2}+(2x-6) \\ \Rightarrow 26={x^2}+4x-6 \\ \Rightarrow {x^2}+4x-32=0 \\ \Rightarrow x=\dfrac{-4\pm \sqrt{{4^2}-4\times -32}}{2} \\ \Rightarrow x=\dfrac{-4\pm \sqrt{16+128}}{2} \\ \Rightarrow x=\dfrac{-4\pm \sqrt{144}}{2} \\ \Rightarrow x=\dfrac{-4\pm 12}{2} \\ \Rightarrow x=-2\pm 6 \\ \therefore x=4,-8 \\ \text{Since age cannot be negative, we have }x = 4. \\ \text{So, the age of eldest child }= 2\times 4=8years $

 Solve for the value of $x : {{\left( {{x}^{2}}+3 \right)}^{2}}-2{{x}^{4}}+2{{x}^{2}}=3$ 

 Start by taking ${{x}^{2}}=a$. Find the value of a using the quadratic formula and then substitute in ${{x}^{2}}=a$ to get the value of $x$.

 $\text{Take }{{x}^{2}}=a ,\text{ then we have } {{\left( a+3 \right)}^{2}}-2{{a}^{2}}+2a=3.\\ \text{On simplifying we get, }{{a}^{2}}-8a-6=0 \\ \text{Using quadratic formula, we have }\\ a=\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times -6}}{2} \\ \Rightarrow a=\dfrac{8\pm \sqrt{88}}{2} \\ \Rightarrow a=4\pm \sqrt{22}.\\ \text{Now, since }{{x}^{2}}=a,\\ \therefore x=\sqrt{4\pm \sqrt{22}} $ 