(D) For equation $I$: $x^{2}-19x+84=0$Splitting the middle term: $x^{2}-7x-12x+84=0$$(x-7)(x-12)=0$So,$x = 7$ or $x = 12$.For equation $II$: $y^{2}-25

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(D) For equation $I$: $x^{2}-19x+84=0$Splitting the middle term: $x^{2}-7x-12x+84=0$$(x-7)(x-12)=0$So,$x = 7$ or $x = 12$.For equation $II$: $y^{2}-25y+156=0$Splitting the middle term: $y^{2}-13y-12y+156=0$$(y-13)(y-12)=0$So,$y = 13$ or $y = 12$.Comparing the values:If $x=7$,then $x If $x=12$,then $x \leq y$ (since $y$ is $12$ or $13$).Combining these,we get $x \leq y$.

Competitive Quantitative Aptitude Algebra QUADRATIC EQUATION

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Solve the given two equations and select the correct option.
$I.$ $x^{2}-19x+84=0$
$II.$ $y^{2}-25y+156=0$

A
if $x > y$
B
if $x \geq y$
C
if $x < y$
D
if $x \leq y$

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Algebra

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QUADRATIC EQUATION

The equation formed by decreasing each root of $ax^2 + bx + c = 0$ by $1$ is $2x^2 + 8x + 2 = 0$. Then:

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If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,the value of $\alpha^{3} + \beta^{3}$ is

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Let $\alpha, \alpha^2$ be the roots of $x^2 + x + 1 = 0$. Then the equation whose roots are $\alpha^{31}, \alpha^{62}$ is:

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Solve the given two equations and select the correct answer from the given options.
$I.$ $12 x^{2} + 11 x - 56 = 0$
$II.$ $4 y^{2} - 15 y + 14 = 0$

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Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

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Solve the given two equations and select the correct option.$I.$ $x^{2}-19x+84=0$$II.$ $y^{2}-25y+156=0$

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The equation formed by decreasing each root of $ax^2 + bx + c = 0$ by $1$ is $2x^2 + 8x + 2 = 0$. Then:

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,the value of $\alpha^{3} + \beta^{3}$ is

Let $\alpha, \alpha^2$ be the roots of $x^2 + x + 1 = 0$. Then the equation whose roots are $\alpha^{31}, \alpha^{62}$ is:

Solve the given two equations and select the correct answer from the given options.$I.$ $12 x^{2} + 11 x - 56 = 0$$II.$ $4 y^{2} - 15 y + 14 = 0$

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

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Solve the given two equations and select the correct option.
$I.$ $x^{2}-19x+84=0$
$II.$ $y^{2}-25y+156=0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $12 x^{2} + 11 x - 56 = 0$
$II.$ $4 y^{2} - 15 y + 14 = 0$